Question

1. AAA reported in 2010 that 24% of people said that they read or sent text messages or email while driving in the previous month. what is the probability that at least 13 of 50 drivers drive and text?

2.If the P-value in a chi-square test is calculated at 0.65, what can you conclude?

a;Not enough information is given. b. the probability of Ha occurring is 65%. c. Reject the null hypothesis. d. Fail to reject the null hypothesis.

3. which of the following is NOT true for a chi-square test?

a. the rule of 2 is a good guideline for determining whether or not to reject the null hypothesis. b. Use only for categorical data.

c. The expected count for each component needs to be at least 5. d. The expected value can be a decimal.

3. Data shoe that 3% of apples produced from an apple orchard are bruised when they reach local store. Compute the probability that at least 2 are bruised in a bushel of 50 apples.

4. we have calculated a confidence interval based on a sample of size n=100. Now we want to get better estimate with a margin of error that only one-fifth as large. How large does our new sample need to be?

5.The American red Cross days that about 45% of the U.S. population has Type O , 40% Type A, 11% Type B, and 4% Type AB. What is the probability that next three donor all have Type A blood?

Answer 1

1. AAA reported in 2010 that 24% of people said that they read or sent text messages or email while driving in the previous month. what is the probability that at least 13 of 50 drivers drive and text?

Probability = p = 0.24

Sample size = n = 50

X: Number of drivers who drive and text.

Here X follows Binomial distribution.

We have to find P(X 13)

P(X 13) = 1 - P(X 12) = 1 - 0.5767 = 0.4233

( Using excel =BINOMDIST(12,50,0.24,1))

The probability that at least 13 of 50 drivers drive and text is 0.4233

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