Math / Data & Statistics

QuestionLunch break: In a recent survey of 655 working Americans ages $25-34$ , the average weekly amount spent on lunch was $\$ 43.88$ with standard deviation $\$ 2.93$ . The weekly amounts are approximately bell-shaped.
Part: 0 / 3
Part 1 of 3 (a) Estimate the percentage of amounts that are between $\$ 35.09$ and $\$ 52.67$ . (Choose one) $\square$ of the amounts fall between $\$ 35.09$ and $\$ 52.67$ . $\square$

Studdy Solution

STEP 1

1. The data is approximately normally distributed (bell-shaped).
2. We can use the empirical rule (68-95-99.7 rule) for normal distributions.
3. The mean ( $\mu$ ) is $\$43.88$ .
4. The standard deviation ( $\sigma$ ) is $\$2.93$ .

STEP 2

1. Calculate the z-scores for $\$35.09$ and $\$52.67$ .
2. Use the z-scores to find corresponding cumulative probabilities.
3. Calculate the percentage of amounts between the two values.

STEP 3

Calculate the z-score for $\$35.09$ using the formula:
$z = \frac{X - \mu}{\sigma}$
where $X = \$35.09$ , $\mu = \$43.88$ , and $\sigma = \$2.93$ .
$z = \frac{35.09 - 43.88}{2.93}$

STEP 4

Simplify the expression to find the z-score for $\$35.09$ .
$z = \frac{-8.79}{2.93} \approx -3.00$

STEP 5

Calculate the z-score for $\$52.67$ using the same formula:
$z = \frac{X - \mu}{\sigma}$
where $X = \$52.67$ , $\mu = \$43.88$ , and $\sigma = \$2.93$ .
$z = \frac{52.67 - 43.88}{2.93}$

STEP 6

Simplify the expression to find the z-score for $\$52.67$ .
$z = \frac{8.79}{2.93} \approx 3.00$

STEP 7

Use the standard normal distribution table to find the cumulative probability for $z = -3.00$ .
$P(Z \leq -3.00) \approx 0.0013$

STEP 8

Use the standard normal distribution table to find the cumulative probability for $z = 3.00$ .
$P(Z \leq 3.00) \approx 0.9987$

STEP 9

Calculate the percentage of amounts between $\$35.09$ and $\$52.67$ by finding the difference between the cumulative probabilities.
$P(-3.00 \leq Z \leq 3.00) = P(Z \leq 3.00) - P(Z \leq -3.00)$ $P(-3.00 \leq Z \leq 3.00) = 0.9987 - 0.0013 = 0.9974$

STEP 10

Convert the decimal to a percentage to estimate the percentage of amounts between $\$35.09$ and $\$52.67$ .
$0.9974 \times 100\% = 99.74\%$
Solution: Approximately $99.74\%$ of the amounts fall between $\$35.09$ and $\$52.67$ .

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Studdy Solution

STEP 1

1. The data is approximately normally distributed (bell-shaped).
2. We can use the empirical rule (68-95-99.7 rule) for normal distributions.
3. The mean ( $\mu$ ) is $\$43.88$ .
4. The standard deviation ( $\sigma$ ) is $\$2.93$ .

STEP 2

1. Calculate the z-scores for $\$35.09$ and $\$52.67$ .
2. Use the z-scores to find corresponding cumulative probabilities.
3. Calculate the percentage of amounts between the two values.

STEP 3

Calculate the z-score for $\$35.09$ using the formula:
$z = \frac{X - \mu}{\sigma}$
where $X = \$35.09$ , $\mu = \$43.88$ , and $\sigma = \$2.93$ .
$z = \frac{35.09 - 43.88}{2.93}$

STEP 4

Simplify the expression to find the z-score for $\$35.09$ .
$z = \frac{-8.79}{2.93} \approx -3.00$

STEP 5

Calculate the z-score for $\$52.67$ using the same formula:
$z = \frac{X - \mu}{\sigma}$
where $X = \$52.67$ , $\mu = \$43.88$ , and $\sigma = \$2.93$ .
$z = \frac{52.67 - 43.88}{2.93}$

STEP 6

Simplify the expression to find the z-score for $\$52.67$ .
$z = \frac{8.79}{2.93} \approx 3.00$

STEP 7

Use the standard normal distribution table to find the cumulative probability for $z = -3.00$ .
$P(Z \leq -3.00) \approx 0.0013$

STEP 8

Use the standard normal distribution table to find the cumulative probability for $z = 3.00$ .
$P(Z \leq 3.00) \approx 0.9987$

STEP 9

Calculate the percentage of amounts between $\$35.09$ and $\$52.67$ by finding the difference between the cumulative probabilities.
$P(-3.00 \leq Z \leq 3.00) = P(Z \leq 3.00) - P(Z \leq -3.00)$ $P(-3.00 \leq Z \leq 3.00) = 0.9987 - 0.0013 = 0.9974$

STEP 10

Convert the decimal to a percentage to estimate the percentage of amounts between $\$35.09$ and $\$52.67$ .
$0.9974 \times 100\% = 99.74\%$
Solution: Approximately $99.74\%$ of the amounts fall between $\$35.09$ and $\$52.67$ .

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Studdy Solution

STEP 1

1. The data is approximately normally distributed (bell-shaped).2. We can use the empirical rule (68-95-99.7 rule) for normal distributions.3. The mean (μ\muμ) is $43.88\$43.88$43.88.4. The standard deviation (σ\sigmaσ) is $2.93\$2.93$2.93.

STEP 2

1. Calculate the z-scores for $35.09\$35.09$35.09 and $52.67\$52.67$52.67.2. Use the z-scores to find corresponding cumulative probabilities.3. Calculate the percentage of amounts between the two values.

STEP 3

Calculate the z-score for $35.09\$35.09$35.09 using the formula:z=X−μσ z = \frac{X - \mu}{\sigma} z=σX−μ​where X=$35.09X = \$35.09X=$35.09, μ=$43.88\mu = \$43.88μ=$43.88, and σ=$2.93\sigma = \$2.93σ=$2.93.z=35.09−43.882.93 z = \frac{35.09 - 43.88}{2.93} z=2.9335.09−43.88​

STEP 4

Simplify the expression to find the z-score for $35.09\$35.09$35.09.z=−8.792.93≈−3.00 z = \frac{-8.79}{2.93} \approx -3.00 z=2.93−8.79​≈−3.00

STEP 5

Calculate the z-score for $52.67\$52.67$52.67 using the same formula:z=X−μσ z = \frac{X - \mu}{\sigma} z=σX−μ​where X=$52.67X = \$52.67X=$52.67, μ=$43.88\mu = \$43.88μ=$43.88, and σ=$2.93\sigma = \$2.93σ=$2.93.z=52.67−43.882.93 z = \frac{52.67 - 43.88}{2.93} z=2.9352.67−43.88​

STEP 6

Simplify the expression to find the z-score for $52.67\$52.67$52.67.z=8.792.93≈3.00 z = \frac{8.79}{2.93} \approx 3.00 z=2.938.79​≈3.00

STEP 7

Use the standard normal distribution table to find the cumulative probability for z=−3.00z = -3.00z=−3.00.P(Z≤−3.00)≈0.0013 P(Z \leq -3.00) \approx 0.0013 P(Z≤−3.00)≈0.0013

STEP 8

Use the standard normal distribution table to find the cumulative probability for z=3.00z = 3.00z=3.00.P(Z≤3.00)≈0.9987 P(Z \leq 3.00) \approx 0.9987 P(Z≤3.00)≈0.9987

STEP 9

STEP 10

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1. The data is approximately normally distributed (bell-shaped).
2. We can use the empirical rule (68-95-99.7 rule) for normal distributions.
3. The mean ( $\mu$ ) is $\$43.88$ .
4. The standard deviation ( $\sigma$ ) is $\$2.93$ .

1. Calculate the z-scores for $\$35.09$ and $\$52.67$ .
2. Use the z-scores to find corresponding cumulative probabilities.
3. Calculate the percentage of amounts between the two values.

Calculate the z-score for $\$35.09$ using the formula:
$z = \frac{X - \mu}{\sigma}$
where $X = \$35.09$ , $\mu = \$43.88$ , and $\sigma = \$2.93$ .
$z = \frac{35.09 - 43.88}{2.93}$

Simplify the expression to find the z-score for $\$35.09$ .
$z = \frac{-8.79}{2.93} \approx -3.00$

Calculate the z-score for $\$52.67$ using the same formula:
$z = \frac{X - \mu}{\sigma}$
where $X = \$52.67$ , $\mu = \$43.88$ , and $\sigma = \$2.93$ .
$z = \frac{52.67 - 43.88}{2.93}$

Simplify the expression to find the z-score for $\$52.67$ .
$z = \frac{8.79}{2.93} \approx 3.00$

Use the standard normal distribution table to find the cumulative probability for $z = -3.00$ .
$P(Z \leq -3.00) \approx 0.0013$

Use the standard normal distribution table to find the cumulative probability for $z = 3.00$ .
$P(Z \leq 3.00) \approx 0.9987$