Data & Statistics

Problem 2201

Encuesta a estudiantes sobre sus preferencias en 3 plataformas: Amazon Prime, HBO y Netflix. Responde:
a.) Estudiantes con las 3 plataformas: b.) Estudiantes con al menos una plataforma: c.) Estudiantes con exactamente una plataforma: d.) Estudiantes con Amazon Prime o HBO, pero no Netflix:

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Problem 2202

Find the mean of the following numbers: 28, 40, 53, 39, 45.

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Problem 2203

Calculate the mean of the data set: 28,40,53,39,4528, 40, 53, 39, 45.

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Problem 2204

Find the median of the following data set: 15, 19, 24, 12, 9, 24, 19, 16, 21, 12.

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Problem 2205

Find the mean of the walking times: 10 (1), 15 (2), 20 (2), 25 (1) minutes. Use frequencies to calculate.

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Problem 2206

Find the median of miles to the nearest airport given the data: 20 (3), 22 (2), 24 (1), 26 (4).

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Problem 2207

A survey of high-school students showed 294 white and 75 minority graduates. What percent are white? Round to one decimal place.

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Problem 2208

Summarize the ethnic distribution of 1772 qualified applicants: %\square \% Black or Hispanic, %\square \% Asian, %\square \% White.

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Problem 2209

A high school surveyed students about post-graduation plans. What percent are white and planning for a 2-year college?

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Problem 2210

A high school survey shows plans of graduates. Calculate the following percentages based on the data provided.

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Problem 2211

A high school survey shows plans of white and minority students. Find percentages for parts a) to c) based on the data.

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Problem 2212

Analyze survey data on high school graduation plans for white and minority students. Calculate various percentages based on the data.

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Problem 2213

Mixing 50 mL50 \mathrm{~mL} of water at 100C100^{\circ} \mathrm{C} with 50 mL50 \mathrm{~mL} at 40C40^{\circ} \mathrm{C} should yield 70C70^{\circ} \mathrm{C}. Why is it 65C65^{\circ} \mathrm{C}? A. Energy lost B. Energy absorbed C. Thermometer error.

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Problem 2214

Find the modal class of wallaby heights and estimate the mean and standard deviation from the given frequency data.

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Problem 2215

Find the modal class of wallaby heights and estimate the mean and standard deviation from the data given.

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Problem 2216

Al surveyed 60 people on restaurant spending.
a) Identify the modal class.
b) Estimate the mean.
c) Estimate the standard deviation and discuss it.
d) Estimate the variance, range, and interquartile range, explaining why they are estimates.

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Problem 2217

Find the interquartile range of the data set: 2, 5, 9, 11, 18, 30, 42, 48, 71, 73, 81. A. 62 B. 21 C. 79 D. 41

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Problem 2218

Miranda scored 6 points in periods 1 & 3, 8 in 2 & 4. Tamera scored 14 in periods 1 & 3. Who scored more points?

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Problem 2219

Miranda scored 6 points in periods 1 & 3, 8 in 2 & 4. Tamera scored 14 in 1 & 3. Which statement is true? F, G, H, or J?

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Problem 2220

Approximate the average rate of change of infections from 0 to 8 days after April 1, 2020, using P(8)5.55P(8) \approx 5.55.

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Problem 2221

Miranda scored 6 points in periods 1 & 3, 8 in 2 & 4. Tamera scored 14 in periods 1 & 3. Which statement is true?

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Problem 2222

Approximate the average rate of change of infections from 0 to 8 days after April 1, 2020, using P(8)P(0)P(8) - P(0).

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Problem 2223

Find the average rate of change of infections from day 0 to day 16, given P(0)=3P(0) = 3, P(8)5.55P(8) \approx 5.55, P(16)10.28P(16) \approx 10.28. Round to two decimal places in thousand people per day.

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Problem 2224

Given 13 college students' credits: 8, 10, 11, 1, 1, 8, 7, 7, 7, 9, 7, 7, 8.
1. Calculate the mean. Discuss its validity.
2. Calculate the median. Discuss its validity.
3. Calculate the mode. Discuss its validity.

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Problem 2225

When 10.7 g10.7 \mathrm{~g} of marble reacts with 60.3 mL60.3 \mathrm{~mL} HCl, how many liters of CO2_2 gas are produced? Density of CO2_2 is 1.798 g/L1.798 \mathrm{~g} / \mathrm{L}.

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Problem 2226

Calculate the extra cost for 2.5 GB used over a 2 GB limit at \0.02perMB:0.02 per MB: st=(0.5 \mathrm{~GB} \times 1000 \mathrm{MB/GB} \times \0.02/MB)0.02/\mathrm{MB}).

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Problem 2227

Tucker sees 20-30 raccoons on clear nights and 10-15 on cloudy nights. What does this suggest about raccoon activity?

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Problem 2228

In a group with a skin disorder, 36 were exposed to methanol, 90 to xylene, and 18 to both. Find those exposed to either.

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Problem 2229

Dibuja un histograma de los tiempos de pago: 19,15,43,39,35,31,27,22,18,14,42,38,13,13,13,41,41,41,37,37,37,3319,15,43,39,35,31,27,22,18,14,42,38,13,13,13,41,41,41,37,37,37,33. Usa límites de clase 12.5 y ancho 7.

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Problem 2230

Encuesta sobre preferencias de streaming: Amazon Prime, HBO y Netflix. Datos: 159 con Amazon, 200 sin Amazon, 192 sin HBO, 137 con Netflix, 108 sin Amazon ni HBO, 104 sin HBO ni Netflix, 124 sin Amazon ni Netflix, 56 sin ninguna.
a.) ¿Cuántos tienen las tres plataformas? b.) ¿Cuántos tienen al menos una plataforma? c.) ¿Cuántos tienen exactamente una plataforma?

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Problem 2231

Find the grouped frequency distribution for these test scores: 83, 85, 97, 91, 92, 82, 90, 89, 91, 83, 93, 88, 86, 84, 98. Class width is 5. Then, draw a frequency polygon with midpoints labeled.

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Problem 2232

Find which two employees have the same hourly rate based on their hours and earnings. Keena's earnings are missing.

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Problem 2233

Find the number of teachers at BCHS teaching Biology, Physical Education, both, or neither using Venn diagrams.

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Problem 2234

Trucks carried peaches: 34 early, 61 late, 50 extra late. Find trucks with only late, only extra late, only one type, and total.

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Problem 2235

How many significant figures are in 8000.0? a) 5 b) 3 c) 2 d) 4

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Problem 2236

Identify a value from the range of function AA given the pairs: (1,1)(-1, 1), (2,2)(-2, 2), (3,3)(-3, 3), (4,4)(-4, 4), (5,5)(-5, 5).

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Problem 2237

Find the average rate of change of YY over the interval [4,4][-4, 4] using the values provided in the table.

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Problem 2238

Create a bar graph for enrollment data: 2006 (2255), 2007 (2363), 2009 (2577), 2010 (2759), 2011 (2605), 2013 (2535), 2014 (2389).

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Problem 2239

Students were surveyed on video game time and test averages. Is the correlation negative or weak? True/False statements provided.

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Problem 2240

Analyze the frequency table of art styles to determine if there's an association between art type and style.

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Problem 2241

A computer's graphics program opens 8 seconds faster and the email program 5 seconds faster. Which has more improvement? Explain.

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Problem 2242

The graphics program improved by -8 seconds and the email program by -5 seconds. Which improved more? Explain.

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Problem 2243

Laura is filling 288 cupcake cups with 66lb66 \mathrm{lb} of batter. Which data types are correct for pounds and cups?

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Problem 2244

Find the standard deviation of ACT scores, mean = 21.5, with 19% above 25. Round to the nearest tenth.

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Problem 2245

A town has a population of 90009000, with 59\frac{5}{9} males. If 40%40\% of males are married, find the married females' percentage.

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Problem 2246

A town has a population of 90009000, with 39\frac{3}{9} males. If 40%40\% of males are married, find the % of married females. (a) 50%50\% (b) 40%40\% (c) 45%45\% (d) None of these

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Problem 2247

When you multiply a value by 100, the decimal moves two places right. Find the percentage of salt in 125 g seasoning with 74.0 g salt.

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Problem 2248

A city inspector found that 41.1%41.1\% of 500 buildings met readiness conditions with a margin of error of ±5%\pm 5\%. What must be true?

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Problem 2249

Graph Jeriel's temperature data starting at 56F56^{\circ} \mathrm{F}, rising and falling over the day, ending at 6060^{\circ}.

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Problem 2250

Sharon Smith evaluates investments X, Y, Z against a 12% return, 6% std. dev. Select investments for risk neutral, averse, and seeking.

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Problem 2251

Solar Designs is evaluating two expansions. Find the return ranges, assess risk, choose an investment, and analyze changes if rates shift.

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Problem 2252

Find the average score of the last four home games: 69,79,58,6669, 79, 58, 66. Did they score more at home or away compared to an average of 61?

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Problem 2253

Solar Designs is evaluating two expansions. Determine the return range for each, assess risk, and choose an investment.

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Problem 2254

Micro-Pub, Inc. is evaluating two cameras (R and S).
a. Find the rate of return range for both cameras. b. Calculate the expected return for each camera. c. Which camera is riskier and why?
Initial investment: \$4,000 for both. Camera R: Pessimistic 20%, Most likely 25% (0.50), Optimistic 30% (0.25). Camera S: Pessimistic 15% (0.20), Most likely 25% (0.55), Optimistic 35% (0.25).

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Problem 2255

What percent of students studied less than 10 minutes if Jeffrey studied for 25 minutes?

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Problem 2256

Find the median for the following group of data items. 70,10,40,10,3070,10,40,10,30
The median is \square (Type an integer or a decimal.)

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Problem 2257

```latex At every point along the AFC curve in the figure to the right, what is true of the dollar amount of this firm's total fixed costs at any given point that one might select?
Total fixed cost is \square by definition. If the output rate is 250 units in the figure, then AFC equals \ \squareperunitsothatTFCequals$ per unit so that TFC equals \$ \square.Itfollowsthenthat,if314unitsareproducedpertimeperiod,AFCmustequal$. It follows then that, if 314 units are produced per time period, AFC must equal \$ \square$ per unit. (Enter your responses rounded to two decimal places.) ```

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Problem 2258

Use the table below to find the percentage of data items in a normal distribution that lie between z=0.7z=0.7 and z=1.6z=1.6. \begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|} \hline z-score & 0.1 & 0.2 & 0.3 & 0.4 & 0.5 & 0.6 & 0.7 & 0.8 & 0.9 & 1.0 \\ Percentile & 53.98 & 57.93 & 61.79 & 65.54 & 69.15 & 72.57 & 75.80 & 78.81 & 81.59 & 84.13 \\ \hline z-score & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6 & 1.7 & 1.8 & 1.9 & 2.0 \\ Percentile & 86.43 & 88.49 & 90.32 & 91.92 & 93.32 & 94.52 & 95.54 & 96.41 & 97.13 & 97.72 \\ \hline z-score & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6 & 2.7 & 2.8 & 2.9 & 3.0 \\ Percentile & 98.21 & 98.61 & 98.93 & 99.18 & 99.38 & 99.53 & 99.65 & 99.74 & 99.81 & 99.87 \\ \hline \end{tabular}
The percentage of data items in a normal distribution that lie between z=0.7z=0.7 and z=1.6z=1.6 is %\% \square

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Problem 2259

Points: 0.94 of 1 Save
According to a certain government traffic safety agency for a large country, the proportion of fatal traffic accidents in the country in which the driver had a positive blood alcohol concentration (BAC) is 0.35 . Suppose a random sample of 104 traffic fatalities in a certain region with a large population results in 44 that involved a positive BAC. Does the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the α=0.1\alpha=0.1 level of significance? Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2).
Start by determining whether the conditions for this hypothesis test are satisfied. Because np0(1p0)=23.7>10\mathrm{np}_{0}\left(1-\mathrm{p}_{0}\right)=23.7>10, the sample size is less than 5%5 \% of the population size, and the sample for testing the hypothesis about the population proportion (Round to one decimal place as needed.) are satisfied. is given to be random, \square the requirements
What are the null and alternative hypotheses? H0\mathrm{H}_{0} : \square \square versus H1\mathrm{H}_{1} : \square >> 0.35 (Type integers or decimals. Do not round.) Find the test statistic, z0z_{0}. z0=z_{0}= \square (Round to two decimal places as needed.)

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Problem 2260

Answer the questions below. (a) The following number of people attended the last 9 screenings of a movie: 117, 195, 197, 198, 199, 202, 204, 207209.117 \text {, 195, 197, 198, 199, 202, 204, } 207 \text {, } 209 .
Which measure should be used to summarize the data? Mean Median Mode (b) Salma has recorded the number of miles she has driven for each of the last 10 weeks: 386,388,392,395,396,402,406,407,415,418.386,388,392,395,396,402,406,407,415,418 .
Which measure should be used to summarize the data? Mean Median Mode (c) Jina wants to determine which letter appears the most often in her favorite poem.
Which measure should she use? Mean Median Mode

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Problem 2261

A group of data items and their mean are given. 21,35,49,84,126,189; Mean =8421,35,49,84,126,189 ; \text { Mean }=84 a. Find the deviation from the mean for each of the data items. b. Find the sum of the deviations in part (a). a. Type the deviation from the mean for each of the data items. \begin{tabular}{|c|c|c|c|c|c|} \hline 21 & 35 & 49 & 84 & 126 & 189 \\ \hline & & & & & \\ \hline \end{tabular}

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Problem 2262

A group of data items and their mean are given. 21,35,49,84,126,189; Mean =8421,35,49,84,126,189 ; \text { Mean }=84 a. Find the deviation from the mean for each of the data items. b. Find the sum of the deviations in part (a). a. Type the deviation from the mean for each of the data items. \begin{tabular}{|c|c|c|c|c|c|} \hline 21 & 35 & 49 & 84 & 126 & 189 \\ \hline-63 & -49 & -35 & 0 & 42 & 105 \\ \hline \end{tabular} b. The sum of the deviations in part (a) is \square

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Problem 2263

Homework \# 4: 7(1,2,3,4)8(2,3,4)7(1,2,3,4) 8(2,3,4) Question 7 of 40 (1 point) | Question Attempt: 1 of 3 1\checkmark 1 ×2\times 2 3\checkmark 3 ×4\times 4 5\equiv 5 6 7 8 9
A normal distribution has mean μ=61\mu=61 and standard deviation σ=20\sigma=20. Find and interpret the zz-score for x=63x=63.
The zz-score for x=63x=63 is \square . So 63 is \square standard deviations (Choose one) the mean μ=61\mu=61.

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Problem 2264

www-awu-aleks.com Secret - Google D... 11/22/24 ATL @ C... Home - Northern... Content ChatGPT Chicago Bulls... Failed to open page Homework * 4: 7(1,2,3,4)8(2,3,4)7(1,2,3,4) 8(2,3,4) Question 8 of 40 (1 point) I Question Attempt 1 of 3 Jonathan 1\checkmark 1 ×2\times 2 3\checkmark 3 ×4\times 4 6 7\checkmark 7 8 9 10 11 12
A normal population has mean μ=7\mu=7 and standard deviation σ=7\sigma=7. Find the probability that a randomly selected value is greater than 4 . Round the answers to at least four decimal places.
The probability that a randomly selected value is greater than 4 is \square .

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Problem 2265

Question 5, 10.1.13 Part 1 of 5 HW Score: 57.14%,457.14 \%, 4 of 7 points Points: 0 of 1 Save
For students who first enrolled in two-year public institutions in a recent semester, the proportion who earned a bachelor's degree within six years was 0.393 . The president of a certain junior college believes that the proportion of students who enroll in her institution have a lower completion rate. (a) State the null and alternative hypotheses in words. (b) State the null and alternative hypotheses symbolically. (c) Explain what it would mean to make a Type I error. (d) Explain what it would mean to make a Type II error. (a) State the null hypothesis in words. Choose the correct answer below. A. Among students who enroll at the certain junior college, the completion rate is less than 0.393. B. Among students who enroll at the certain junior college, the completion rate is greater than 0.393. C. Among students who enroll at the certain junior college, the completion rate is 0.393 . D. Among students who first enroll in two-year public institutions, the completion rate is 0.393 .

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Problem 2266

Secret - Google D... www-awu.aleks.com Homework \# 4: 7(1,2,3,4)8(2,3,4)7(1,2,3,4) 8(2,3,4) Home - Northern... Content ChatGPT Chicago Bulls... Failed to open page Question 9 of 40 (1 point) I Question Attempt: 1 of 3 Jonathan 1\checkmark 1 ×2\times 2 3\checkmark 3 =5=5 6 67 9 10 11 12
A normal population has mean μ=37\mu=37 and standard deviation σ=14\sigma=14. Find the value that has 25%25 \% of the population above it. Round the answer to at least one decimal place.
The value that has 25%25 \% of the population above it is \square .

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Problem 2267

www-awu.aleks.com Betting an... Secret - Google D... 11/22/24 ATL @ C... Home - Northern... Content ALEKS - Jonathan... ChatGPT Chic Homework \# 4: 7(1,2,3,4)8(2,3,4)7(1,2,3,4) 8(2,3,4) Question 11 of 40 (1 point) I Question Attempt: 1 of 3 1\checkmark 1 ×2\times 2 3\checkmark 3 4 6 7\checkmark 7 \checkmark 9\checkmark 9
A normal population has mean μ=9\mu=9 and standard deviation σ=5\sigma=5. (a) What proportion of the population is less than 20? (b) What is the probability that a randomly chosen value will be greater than 6 ?
Round the answers to four decimal places.
Part: 0/20 / 2
Part 1 of 2
The proportion of the population less than 20 is \square . Skip Part Check - 2024 McGraw Hill LLC. All Ri

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Problem 2268

www-awu.aleks.com Secret - Google D... 11/22/24 ATL @ C... Home - Northern... Content ChatGPT Chicas Homework \# 4: 7(1,2,3,4) 8(2,3,4) Question 11 of 40 (1 point) I Question Attempt: 1 of 3 1\checkmark 1 3\checkmark 3 4 6 7 8 9
A normal population has mean μ=9\mu=9 and standard deviation σ=5\sigma=5. (a) What proportion of the population is less than 20? (b) What is the probability that a randomly chosen value will be greater than 6 ?
Round the answers to four decimal places.
Part 1 of 2
The proportion of the population less than 20 is 0.9861 .
Part: 1 / 2
Part 2 of 2
The probability that a randomly chosen value will be greater than 6 is \square .

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Problem 2269

1 Matching 5 points
Determine if each of the following situations is a binomial distribution. If it does, explain why each condition is met. If it does not, say what condition was not met
I roll a die 100 times and count how many of each number I get.
I flip a coin and count how many times it takes to land on a "head".
I spin a spinner 20 times and count how many times it lands on blue.
I flip a coin 12 times and count how many heads I get. I throw a bean bag at a cornhole board 10 times. I receive coaching after each throw. I count how many bags go in.

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Problem 2270

Secret - Google D... 11/22/24 ATL @ C... www-awu.aleks.com Homework \# 4: 7(1,2,3,4)8(2,3,4)7(1,2,3,4) 8(2,3,4) Question 13 of 40 (1 point) I Question Attempt: 1 of 3 ALEKS - Jonathan... ChatGPT Chicago Bulls... Failec 1\checkmark 1 3\checkmark 3 ×4\times 4 6 7\checkmark 7 8\checkmark 8 9 10 11\checkmark 11
A sample of size 190 will be drawn from a population with mean 45 and standard deviation 11. Use the TI-83 Plus/TI-84 Plus calculator. Part 1 of 2 (a) Find the probability that xˉ\bar{x} will be less than 43 . Round the answer to at least four decimal places.
The probability that xˉ\bar{x} will be less than 43 is 0.0060 .
Part: 1/21 / 2
Part 2 of 2 (b) Find the 60th 60^{\text {th }} percentile of xˉ\bar{x}. Round the answer to at least two decimal places.
The 60th 60^{\text {th }} percentile is \square. Skip Part Check

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Problem 2271

www-awu.aleks.com s Betting an.... Secret - Google D... Home - Northern... Content ChatGPT Chicago Bulls... Failed to open page Homework \& 4: 7(1,2,3,4)3(2,3,4)7(1,2,3,4) 3(2,3,4) Question 14 of 40 (1 point) I Question Attempt: 1 of 3 Jonathan 1\checkmark 1 3\checkmark 3 ×4\times 4 6 7\checkmark 7 8\checkmark 8 9\checkmark 9 10\checkmark 10 11\checkmark 11 Español
Watch your cholesterol: The mean serum cholesterol level for U.S. adults was 198, with a standard deviation of 39.3 (the units are milligrams per deciliter). A simple random sample of 106 adults is chosen. Use the TI-84 Plus calculator. Round the answers to at least four decimal places.
Part 1 of 3 (a) What is the probability that the sample mean cholesterol level is greater than 205?
The probability that the sample mean cholesterol level is greater than 205 is 0.0331 .
Part 2 of 3 (b) What is the probability that the sample mean cholesterol level is between 186 and 192?
The probability that the sample mean cholesterol level is between 186 and 192 is 0.0572 .
Part: 2/32 / 3
Part 3 of 3 (c) Using a cutoff of 0.05 , would it be unusual for the sample mean to be less than 194?
It \square (Choose one) be unusual for the sample mean to be less than 194, since the probability is \square . Skip Part Check Save For Later Submit Assignmen

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Problem 2272

A random sample of students was surveyed and asked to list their grade level and what movie genre they prefer. Results are shown in the table below.
Movie Genre \begin{tabular}{|c|c|c|c|c|} \hline & Superhero & Comedy & Drama & Total \\ \hline 6th grade & 11 & 15 & 10 & 36 \\ \hline 7th Grade & 16 & 19 & 14 & 49 \\ \hline 8th Grade & 14 & 20 & 16 & 50 \\ \hline Total & 41 & 54 & 40 & 135 \\ \hline \end{tabular}
What percent of the students prefer superhero movies? Round your answer to the nearest tenth of a percent. Answer Attempt 1 out of 5 Nov 22 8:51 US

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Problem 2273

Secret - Google D. www-awy.aleks.com igan. 11/22/24 ATL @ C. C. Home - Northern... A aleks - Jonathan... ChatGPT Chicago Bulls... Falied to open page Homework \& 4: 7(1,2,3,4)8(2,3,4)7(1,2,3,4) 8(2,3,4) Question 15 of 40 (1 point) I Question Attempt: 1 of 3 1\checkmark 1 ×2\times 2 - 3 ×4\times 4 =5=5 6 7\checkmark 7 \checkmark 9 10\checkmark 10 11\checkmark 11 Jonathan
TV sets: According to the Nielsen Company, the mean number of TV sets in a U.S. household was 2.24. Assume the standard deviation is 1.2 . A sample of 95 households is drawn.
Part: 0/50 / 5 \square Part 1 of 5 (a) What is the probability that the sample mean number of TV sets is greater than 2? Round your answer to at least four decimal places.
The probability that the sample mean number of TV sets is greater than 2 is \square . Skip Part Check Save For Later Submit Assignment - 2024 MaGraw Hill LLC. All Rights Reserved. Terms of Use I Pivacy Center I Accessiblity

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Problem 2274

1\checkmark 1 2 3\checkmark 3 6 7 8 10 11\checkmark 11 12 Español
TV sets: According to the Nielsen Company, the mean number of TV sets in a U.S. household was 2.24. Assume the standard deviation is 1.2 . A sample of 95 households is drawn.
Part 1 of 5 (a) What is the probability that the sample mean number of TV sets is greater than 2? Round your answer to at least four decimal places.
The probability that the sample mean number of TV sets is greater than 2 is 0.9744 .
Part: 1/51 / 5
Part 2 of 5 (b) What is the probability that the sample mean number of TV sets is between 2.5 and 3? Round your answer to at least four decimal places.
The probability that the sample mean number of TV sets is between 2.5 and 3 is \square . Skip Part Check Save For Later Submit Assignment

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Problem 2275

There are 48 states in a certain country. The histogram shows murder rates per 100,000 residents and the number of states that had these rates in a certain year. Use this information to complete parts (a) through (e) to the right.
Murder Rates per 100,000 Residents, by State a. Is the shape of this distribution best classified as normal, skewed left, or skewed right?
Choose the correct distribution type below. A. Normal distribution B. Skewed left c. Skewed right b. Uniform distribution b. Calculate the mean murder rate per 100,000 residents for the states.
The mean murder rate is \square murders per 100,000 residents. (Type an integer or decimal rounded to two decimal place as needed.)

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Problem 2276

(0) Watch Video Show Examples each week (for of surveyed as to how much non-school screen time they had media, watching TV, or playing vivey, screen time was defined as: time spent online, on social
Screen Time \begin{tabular}{|c|c|c|c|} \hline & above & below & Total \\ \hline less than 4\mathbf{4} hours & 8 & 6 & 14 \\ \hline 48\mathbf{4 - 8} hours & 18 & 16 & 34 \\ \hline 812\mathbf{8 - 1 2} hours & 19 & 16 & 35 \\ \hline more than 12 hours & 13 & 20 & 33 \\ \hline Total & 58 & 58 & 116 \\ \hline \end{tabular}
What percent of the students who spend 4-8 hours a week on screens reported a grade average above 80 ? Round your answer to the nearest tenth of a percent.

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Problem 2277

Watch Video Show Examples
A survey stopped men and women at random to ask them where they purchased groceries, at a local grocery store or online.
Grocery Options \begin{tabular}{|c|c|c|c|} \hline & Store & Online & Total \\ \hline Women & 37 & 12 & 49 \\ \hline Men & 40 & 11 & 51 \\ \hline Total & 77 & 23 & 100 \\ \hline \end{tabular}
What percent of the men surveyed shop online? Round your answer to the nearest whole number pércent.

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Problem 2278

Question Watch Video Show Examples
A survey asked a group of adults and youths if they prefer reading books printed on paper or electronic books.
Book Preference \begin{tabular}{|c|c|c|c|} \hline & Print & Electronic & Total \\ \hline Youths & 27 & 40 & 67 \\ \hline Adults & 29 & 27 & 56 \\ \hline Total & 56 & 67 & 123 \\ \hline \end{tabular}
What percent of the adults surveyed prefer reading electronic books? Round your answer to the nearest tenth of a percent. Answer Attempt 1 out of 5

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Problem 2279

The scores on a test are normally distributed with a mean of 200 and a standard deviation of 10 . Find the score that is 3123 \frac{1}{2} standard deviations above the mean.
A score of \square is 3123 \frac{1}{2} standard deviations above the mean.

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Problem 2280

Find the range for the group of data items. 22,23,24,25,2622,23,24,25,26
The range is \square

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Problem 2281

A group of data items and their mean are given. 21,35,49,84,126,189; Mean =8421,35,49,84,126,189 ; \text { Mean }=84 a. Find the deviation from the mean for each of the data items. b. Find the sum of the deviations in part (a). a. Type the deviation from the mean for each of the data items. \begin{tabular}{|c|c|c|c|c|c|} \hline 21 & 35 & 49 & 84 & 126 & 189 \\ \hline & & & & & \\ \hline \end{tabular} b. The sum of the deviations in part (a) is \square

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Problem 2282

Changing jobs: A sociologist sampled 214 people who work in computer-related jobs, and found that 44 of them have changed jobs in the past 6 months.
Part: 0/20 / 2
Part 1 of 2 (a) Construct a 99.8%99.8 \% confidence interval for those who work in computer-related jobs who have changed jobs in the past 6 months. Round the answer to at least three decimal places. A 99.8%99.8 \% confidence interval for the proportion of those who work in computer-related jobs who have changed jobs in the past 6 months is \square <p<<p< \square .

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Problem 2283

This question: 1 point(s) possible Subm
Find the mean for the data items in the given frequency distribution. \begin{tabular}{|l|l|l|l|l|l|l|l|l|} \hline Score, x\mathbf{x} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline Frequency, f\mathbf{f} & 3 & 6 & 2 & 5 & 3 & 1 & 3 & 2 \\ \hline \end{tabular}
The mean is \square (Round to 3 decimal places as needed.)

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Problem 2284

Find the critical values for a 95%95 \% confidence interval using the chi-square distribution with 20 degrees of freedom. Round the answers to three decimal places.
The critical values are \square and \square .

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Problem 2285

Construct a 95%95 \% confidence interval for the population standard deviation σ\sigma if a sample of size 17 has standard deviation s=6.7s=6.7. Round the answers to at least two decimal places.
A 95%95 \% confidence interval for the population standard deviation is: \square \square <σ<<\sigma<

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Problem 2286

IQ scores: Scores on an IQ test are normally distributed. A sample of 8 IQ scores had standard deviation s=6s=6. (a) Construct a 95%95 \% confidence interval for the population standard deviation σ\sigma. Round the answers to at least two decimal places. (b) The developer of the test claims that the population standard deviation is σ=7\sigma=7. Does this confidence interval contradict this claim? Explain.
Part: 0/20 / 2
Part 1 of 2
A 95%95 \% confidence interval for the population standard deviation is \square <σ<<\sigma< \square .

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Problem 2287

Points: 0 of 1 Save
The standard deviation in the pressure required to open a certain valve is known to be σ=1,4psi\sigma=1,4 \mathrm{psi}. Due to changes in the manufacturing process, the quality-control manager feels that the pressure variability has increased. (a) State the null and alternative hypotheses in words. (b) State the null and alternative hypotheses symbolically. (c) Explain what it would mean to make a Type I error. (d) Explain what it would mean to make a Type II error. B. The standard deviation in the pressure required to open a certain valve is greater than 1.4 psi. C. The standard deviation in the pressure required to open a certain valve is different from 1.4 psi . D. The standard deviation in the pressure required to open a certain valve is 1.4 psi .
State the alternative hypothesis in words. Choose the correct answer below. A. The standard deviation in the pressure required to open a certain valve is different from 1.4 psi. B. The standard deviation in the pressure required to open a certain valve is 1.4 psi . C. The standard deviation in the pressure required to open any valve is 1.4 psi . D. The standard deviation in the pressure required to open a certain valve is greater than 1.4 psi . (b) State the hypotheses symbolically. H0\mathrm{H}_{0} : \square == 1.4 psi H1H_{1} : σ\sigma >> 1.4 psi (Type integers or decimals. Do not round.) (c) What would it mean to make a Type I error?
The manager \square the hypothesis that the pressure variability is \square \square psi , when the true pressure variability is \square \square psi. (Type integers or decimals. Do not round.)

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Problem 2288

Calculate ΔGrxn for the following reaction at 298 K. Use the ΔGf values in this table of thermodynamic properties.\text{Calculate } \Delta G_{\mathrm{rxn}} \text{ for the following reaction at 298 K. Use the } \Delta G_{\mathrm{f}}^{\circ} \text{ values in this table of thermodynamic properties.}
Cr2O3(s)+3CO(g)2Cr(s)+3CO2(g)\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s}) + 3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Cr}(\mathrm{s}) + 3 \mathrm{CO}_{2}(\mathrm{g})
ΔGrxn=\Delta G_{\mathrm{rxn}}^{\circ} = \square
 kJ\square \text{ kJ}
\text{Is this reaction spontaneous or nonspontaneous at 298 K?}
\text{nonspontaneous}
\text{spontaneous}
\text{The standard Gibbs free energy of formation values are:}
ΔGf for Cr2O3(s) is 1058.1 kJ/mol\Delta G_{\mathrm{f}}^{\circ} \text{ for } \mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s}) \text{ is } -1058.1 \text{ kJ/mol}
ΔGf for CO(g) is 137.2 kJ/mol\Delta G_{\mathrm{f}}^{\circ} \text{ for } \mathrm{CO}(\mathrm{g}) \text{ is } -137.2 \text{ kJ/mol}
ΔGf for Cr(s) is 0 kJ/mol\Delta G_{\mathrm{f}}^{\circ} \text{ for } \mathrm{Cr}(\mathrm{s}) \text{ is } 0 \text{ kJ/mol}
ΔGf for CO2(g) is 394.4 kJ/mol\Delta G_{\mathrm{f}}^{\circ} \text{ for } \mathrm{CO}_{2}(\mathrm{g}) \text{ is } -394.4 \text{ kJ/mol}

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Problem 2289

11 12\checkmark 12 13 15 16 17 18 19 20
Coffee: The National Coffee Association reported that 65%65 \% of U.S. adults drink coffee daily. A random sample of 300U.S300 \mathrm{U} . \mathrm{S}. adults is selected. Round your answers to at least four decimal places as needed.
Part 1 of 6 (a) Find the mean μp\mu_{p}.
The mean μp^\mu_{\hat{p}} is 0.65 .
Part 2 of 6 (b) Find the standard deviation σp^\sigma_{\hat{p}}.
The standard deviation σp^\sigma_{\hat{p}} is 0.0275 .
Part 3 of 6 (c) Find the probability that more than 66%66 \% of the sampled adults drink coffee daily.
The probability that more than 66%66 \% of the sampled adults drink coffee daily is 0.3564 .
Part 4 of 6 (d) Find the probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71 .
The probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71 is 0.9836 . Part 5 of 6 Cava For 1 ater Submi (e) Find the probability that less than 60%60 \% of sampled adults drink coffee daily.
The probability that less than 60%60 \% of sampled aduts drink coffee đaity is \square .

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Problem 2290

11 12 13
Coffee: The National Coffee Association reported that 65%65 \% of U.S. adults drink coffee daily. A random sample of 300 U.S. adults is selected. Round your answers to at least four decimal places as needed.
Part 1 of 6 (a) Find the mean μp\mu_{p}.
The mean μp^\mu_{\hat{p}} is 0.65 .
Part 2 of 6 (b) Find the standard deviation σp^\sigma_{\hat{p}}.
The standard deviation σp^\sigma_{\hat{p}} is 0.0275 .
Part 3 of 6 (c) Find the probability that more than 66%66 \% of the sampled adults drink coffee daily.
The probability that more than 66%66 \% of the sampled adults drink coffee daily is 0.3564 .
Part 4 of 6 (d) Find the probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71 .
The probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71 is 0.9836 . Part: 5/65 / 6
Part 6 of 6 (f) Using a cutoff of 0.05 , would it be unusual if less than 63%63 \% of the sampled adults drink coffee daily? it \square (Choose one) V\mathbf{V} be unusual if less than 63%63 \% of the sampled adults drink coffee daily, since the probability is \square would would not

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Problem 2291

Coffee: The National Coffee Association reported that 63%63 \% of U.S. adults drink coffee daily. A random sample of 250 U.S. adults is selected. Round your answers to at least four decimal places as needed.
Part 1 of 2 (a) Find the probability that more than 64%64 \% of the sampled adults drink coffee daily.
The probability that more than 64%64 \% of the sampled adults drink coffee daily is 0.3721 .
Part: 1/21 / 2
Part 2 of 2 (b) Find the probability that the proportion of the sampled adults who drink coffee daily is between 0.62 and 0.68 .
The probability that the proportion of the sampled adults who drink coffee daily is between 0.62 and 0.68 is \square.

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Problem 2292

Homework \# 4: 7(1,2,3,4)8(2,3,4)7(1,2,3,4) 8(2,3,4) Question 21 of 40 (1 point) I Question Attempt: 1 of 3 Jonatha 12\checkmark 12 13 14\checkmark 14 15 16 18 19 20 21\equiv 21 22\checkmark 22
Student loans: The Institute for College Access and Success reported that 65%65 \% of college students in a recent year graduated with student loan debt. A random sample of 90 graduates is drawn. Round your answers to at least four decimal places if necessary.
Part 1 of 6 (a) Find the mean μp^\mu_{\hat{p}}.
The mean μp\mu_{p} is 0.65 .
Part: 1 / 6
Part 2 of 6 (b) Find the standard deviation σp^\sigma \hat{p}.
The standard deviation σp^\sigma \hat{p} is \square Save For Later Subm

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Problem 2293

acret - Coogle... 11/22/24 ATL @... Home - Norther... Content ChatGPT Chicago Bull.. Failed to open... Homework \# 4: 7(1,2,3,4)8(2,3,4)7(1,2,3,4) 8(2,3,4) Question 21 of 40 (1 point) I Question Attempt: 1 of 3 12\checkmark 12 13 14\checkmark 14 15 16\equiv 16 =18=18 19 20 21\equiv 21
Student loans: The Institute for College Access and Success reported that 65%65 \% of college students in a recent year graduated with student loan debt. A random sample of 90 graduates is drawn. Round your answers to at least four decimal places if necessary.
Part 1 of 6 (a) Find the mean μp^\mu_{\hat{p}}.
The mean μp^\mu_{\hat{p}} is 0.65 .
Part 2 of 6 (b) Find the standard deviation σp^\sigma \hat{p}.
The standard deviation σp^\sigma_{\hat{p}} is 0.0503 . pp
Part: 2/62 / 6
Part 3 of 6 (c) Find the probability that less than 52%52 \% of the people in the sample were in debt.
The probability that less than 52%52 \% of the people in the sample were in debt is \square . Skip Part Check Save For Later Submit A Q 2024 McGraw Hill LLC All Rights Reserved. Terms of Uso - Fivacy Center

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Problem 2294

\text{Student loans: The Institute for College Access and Success reported that } 65\% \text{ of college students in a recent year graduated with student loan random sample of 90 graduates is drawn. Round your answers to at least four decimal places if necessary.}
\text{Part 1 of 6} \begin{itemize} \item[(a)] \text{Find the mean } \mu_{\hat{p}}. \begin{itemize} \item \text{The mean } \mu_{\hat{p}} \text{ is } 0.65. \end{itemize} \end{itemize}
\text{Part 2 of 6} \begin{itemize} \item[(b)] \text{Find the standard deviation } \sigma. \begin{itemize} \item \text{The standard deviation } \sigma_{\hat{p}} \text{ is } 0.0503. \end{itemize} \end{itemize}
\text{Part 3 of 6} \begin{itemize} \item[(c)] \text{Find the probability that less than } 52\% \text{ of the people in the sample were in debt.} \begin{itemize} \item \text{The probability that less than } 52\% \text{ of the people in the sample were in debt is } 0.0049. \end{itemize} \end{itemize}
\text{Part 4 of 6} \begin{itemize} \item[(d)] \text{Find the probability that between } 60\% \text{ and } 80\% \text{ of the people in the sample were in debt.} \begin{itemize} \item \text{The probability that between } 60\% \text{ and } 80\% \text{ of the people in the sample were in debt is } 0.8375. \end{itemize} \end{itemize}
\text{Part 5 of 6} \begin{itemize} \item[(e)] \text{Find the probability that more than } 70\% \text{ of the people in the sample were in debt.} \begin{itemize} \item \text{The probability that more than } 70\% \text{ of the people in the sample were in debt is } \square. \end{itemize} \end{itemize}

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Problem 2295

\text{Student loans: The Institute for College Access and Success reported that } 65\% \text{ of college students in a recent year graduated with student loan debt. A random sample of 90 graduates is drawn. Round your answers to at least four decimal places if necessary.}
\text{Part 1 of 6} \text{(a) Find the mean } \mu_{\hat{p}}.
\text{The mean } \mu_{\hat{p}} \text{ is } 0.65.
\text{Part 2 of 6} \text{(b) Find the standard deviation } \sigma_{\hat{p}}.
\text{Part 6 of 6} \text{(f) Using a cutoff of } 0.05, \text{ would it be unusual if less than } 66\% \text{ of people in the sample were in debt?}
\text{It (Choose one) } \nabla \text{ be unusual if less than } 66\% \text{ of the people in the sample were in debt, since the probability is } \square.

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Problem 2296

21 22 23 24 25 26 28 29 30 31
Fake Twitter followers: Many celebrities and public figures have Twitter accounts with large numbers of followers. However, some of these followers are fake, resulting from accounts generated by spamming computers. In a sample of 64 twitter audits, the mean percentage of fake followers was 14.6 with a standard deviation of 10.6 .
Part 1 of 2 (a) Construct a 99.5%99.5 \% confidence interval for the mean percentage of fake Twitter followers. Round the answers to at least one decimal place,
A 99.5%99.5 \% confidence interval for the mean percentage of fake Twitter followers is 10.7<μ<18.510.7<\mu<18.5.
Part: 1/21 / 2
Part 2 of 2 (b) Based on the confidence interval, is it reasonable that the mean percentage of fake Twitter followers is less than 9?
It (Choose one) \boldsymbol{\nabla} reasonable that the mean percentage of fake Twitter followers is less than 9 . Save For Later Submit Assignme Check - 2024 McGraw Hill LLC. All Rights Reserved. Terms of Use| Privacy Center| Accessi

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Problem 2297

www-awu.aleks.com s Bettin Secret - Goo... Home - Nort... Content ChatGPT Chicago B. Failed to ope. Oniline Bertin. (5) KaCe Homework \& 4: 7(1,2,3,4)8(2,3,4)7(1,2,3,4) 8(2,3,4) Question 28 of 40 (1 point) I Question Attempt: 1 of 3 21 22 23 25 26 27 =28=28 29 30 Jonathan
Baby weights: Following are weights, in pounds, of 10 two-month-old baby girls. It is reasonable to assume that the population is approximately normal. \begin{tabular}{ccccc} \hline 12.32 & 11.87 & 12.34 & 11.48 & 12.66 \\ 8.51 & 14.13 & 12.95 & 9.34 & 8.63 \\ \hline \end{tabular} Send data to Excel (a) Construct a 90%90 \% confidence interval for the mean weight of two-month-old baby girls. (b) According to the National Health Statistics Reports, the mean weight of two-month-old baby boys is 13.9 pounds. Based on the confidence interval, is it reasonable to believe that the mean weight of two-month-old baby girls may be the same as that of two-month-old baby boys? Explain.
Part 1 of 2 (a) A 90%90 \% confidence interval for the mean weight of two-month-old baby girls is 10.307<μ<12.53910.307<\mu<12.539.
Part: 1/21 / 2
Part 2 of 2 (b) It (Choose one) \boldsymbol{\nabla} reasonable to believe that the mean weight of two-month-old baby girls may be the same as that of two-month-old baby boys. Skip Part Check Save For Later Submit Assi - 2024 McGraw Hill LLC. All Rights Reserved. Terms of Use 1 Pivacy Center \quad t

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Problem 2298

is Bettin. Secret - Goo... ww-awu.aleks.com Homework * 4:7(1,2,3,4) 8(2,3,4) Question 30 of 40 (1 point) I Question Attempt: 1 of 3 Chatept Chicago B. Failed to opo... Online Boting. (5) KalCe... Jonathan 2 23 24 25 26 27 28 29 30 31 32
Smart phone: Among 240 cell phone owners aged 18-24 surveyed, 108 said their phone was an Android phone. Perform the following.
Part: 0/30 / 3
Part 1 of 3 (a) Find a point estimate for the proportion of cell phone owners aged 18-24 who have an Android phone. Round the answer to at least three decimal places.
The point estimate for the proportion of cell phone owners aged 18 - 24 who have an Android phone is \square . Skip Part Check Save For Later Submit Assi - 2024 MaGraw Hill LLC. All Righis Resenced, Terms of Use 1 Pinasy Center

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Problem 2299

www-awu.aleks.com Secret - GoO. Home - Nort Content ChatGPT Chicago B.. Failed to ope... Online Bettin... (5) KaiCe... Homework \4:7(1,2,3,4)8(2,3,4)Question30of40(1point)IQuestionAttempt:1of3Jonathan 4: 7(1,2,3,4) 8(2,3,4) Question 30 of 40 (1 point) I Question Attempt: 1 of 3 Jonathan \checkmark 23242526272829 24 25 26 27 28 29 \equiv 30$ 31 32 Esparí 33
Smart phone: Among 240 cell phone owners aged 18-24 surveyed, 108 said their phone was an Android phone. Perform the following.
Part 1 of 3 (a) Find a point estimate for the proportion of cell phone owners aged 18-24 who have àn Android phone. Round the answer to at least three decimal places.
The point estimate for the proportion of cell phone owners aged 18 - 24 who have an Android phone is 0.450 .
Part: 1/31 / 3
Part 2 of 3 (b) Construct a 90%90 \% confidence interval for the proportion of cell phone owners aged 18-24 who have an Android phone. Round the answer to at least three decimal places.
A 90%90 \% confidence interval for the proportion of cell phone owners aged 18-24 who have an Android phone is \square <p<<p< \square . Skip Part Check Save For Later Submit Assignr Q 2024 McGraw Hill LLC. All Rights Reserved. Terms of Use I Privacy Center I Acce

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Problem 2300

www-awuialeks.com Secret - Coo... 11/22/24 ATL... Home - Nort... Content ChatGPT Chicago B... Failed to ope... Online Bettin... (6) KaiCe. Homework * 4: 7(1,2,3,4)8(2,3,4)7(1,2,3,4) 8(2,3,4) Question 30 of 40 (1 point) I Question Attempt: 1 of 3 Jonath
Smart phone: Among 240 cell phone owners aged 18-24 surveyed, 108 said their phone was an Android phone. Perform the following.
Part 1 of 3 (a) Find a point estimate for the proportion of cell phone owners aged 18-24 who have an Android phone. Round the answer to at least three decimal places.
The point estimate for the proportion of cell phone owners aged 182418-24 who have android phone is 0.450 .
Part 2 of 3 (b) Construct a 90%90 \% confidence interval for the proportion of cell phone owners aged 18-24 who have an Android phone. Round the answer to at least three decimal places.
A 90%90 \% confidence interval for the proportion of cell phone owners aged 18-24 who have an Android phone is 0.397<p<0.5030.397<p<0.503
Part: 2/32 / 3
Part 3 of 3 (c) Assume that an advertisement claimed that 38%38 \% of cell phone owners aged 18-24 have an Android phone. Does the confidence interval contradict this claim? claim?
The confidence interval \square (Choose one) contradict the claim, because 0.38 (Choose one) contained in the confidence interval. Skip Part Check Save For Later Submit Assignm ( 2024 MaGraw Hill LLC. All Rights Reserved. Terms of Use I Privacy Center I Acces:

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