Math  /  Data & Statistics

Question15. Records show that the probability of seeing a hawk migrating on a day in September is about 35%35 \%. What is the minimum number of days a person must watch to be at least 96.8%96.8 \% sure of seeing one or more hawks migrating? a) 5 b) 6 c) 7 d) 8 e) 9
Use the table below for numbers 16 and 17. The following data are based on a survey taken by a consumer research firm. In this table. x=x= number of televisions in a household and %=\%= percentages of U.S. households. \begin{tabular}{|l|l|l|l|l|l|l|} \hline x\boldsymbol{x} & 0 & 1 & 2 & 3 & 4 & 5 or more \\ \hline%\% & 3%3 \% & 11%11 \% & 28%28 \% & 39%39 \% & 12%12 \% & 7%7 \% \\ \hline \end{tabular}
16. What is the probability that a household selected at random has less than three televisions? a) 0.81 b) 0.39 c) 0.42 d) 0.58 e) 0.19
17. The probability that a manufacture part at a plant is defective is 0.02\mathbf{0 . 0 2}. The plant has manufactured 300 parts. Let rr be the random variable representing the number of defective parts. What is the probability that exactly five parts are defectlve? Use the Poisson distribution for the binomial. a) 0.0268 b) 0.9732 c) 0.8394 d) 0.001 e) 0.1606
18. Compute the expected value of the xx distribution (round televisions of five or more to five). a) 15 b) 2.67 c) 1.28 d) 1.13 e) 3.1
19. The number of points scored in a soccer game is an example of what type of variable? a) Discrete b) Continuous c) Both A and B

20r. A business has seven locations to choose from and wishes to rank only the top three locations. How many different ways can this be done? a) 5,040 b) 840 c) 420 d) 210 e) 35
21r. Suppose a researcher assigns the numbers to possible study participants and then chooses every 10th 10^{\text {th }} subject after randomly selecting a starting value. This is an example of what type of sampling? a) Random b) Systematic c) Stratified d) Cluster e) Convenience

Studdy Solution

STEP 1

What is this asking? How many days in September do we need to watch to be super sure (96.8%96.8\%) we'll see at least one hawk migrating? Watch out! Don't mix up "at least one hawk" with "exactly one hawk".
We're looking for the probability of seeing *one or more* hawks.

STEP 2

1. Flip the Script
2. Calculate the Probability
3. Find the Number of Days

STEP 3

Instead of figuring out the probability of seeing *at least one* hawk, let's think about the opposite: the probability of *not* seeing *any* hawks!
This makes the math way easier.
If we're 96.8%96.8\% sure of seeing at least one hawk, then we're 100%96.8%=3.2%100\% - 96.8\% = 3.2\% sure of *not* seeing any hawks.
So, we want P(no hawks)=0.032P(\text{no hawks}) = 0.032.

STEP 4

We know the probability of seeing a hawk on any given day is 35%35\%, which means P(hawk)=0.35P(\text{hawk}) = 0.35.
Therefore, the probability of *not* seeing a hawk on any given day is 10.35=0.651 - 0.35 = 0.65, so P(no hawk)=0.65P(\text{no hawk}) = 0.65.

STEP 5

Let 'n' be the number of days we watch.
The probability of not seeing any hawks for 'n' days is P(no hawks for n days)=(0.65)nP(\text{no hawks for n days}) = (0.65)^n.
We want this to be equal to 0.0320.032.
So, we have the equation (0.65)n=0.032(0.65)^n = 0.032.

STEP 6

To solve for 'n', we can use logarithms.
Taking the logarithm of both sides (we'll use base 10 here, but any base would work!), we get log((0.65)n)=log(0.032)\log((0.65)^n) = \log(0.032).
Using the logarithm power rule, this becomes nlog(0.65)=log(0.032)n \cdot \log(0.65) = \log(0.032).

STEP 7

Now, we can isolate 'n' by dividing both sides by log(0.65)\log(0.65): n=log(0.032)log(0.65) n = \frac{\log(0.032)}{\log(0.65)} n1.494850.18709 n \approx \frac{-1.49485}{-0.18709} n7.99 n \approx 7.99

STEP 8

Since 'n' must be a whole number of days, we round up to the nearest whole number, which is **8**.
We can't watch for a fraction of a day!

STEP 9

We need to watch for **8** days to be at least 96.8%96.8\% sure of seeing at least one hawk migrating.
The answer is (d).

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