Question

Suppose the time between buses at a particular stop is a positively skewed random variable with an average of 59 minutes and standard deviation of 6 minutes. Suppose the time between buses at this stop is measured for a randomly selected week, resulting in a random sample of n = 38 times. The average of this sample, X , is a random variable that comes from a specific probability distribution. (a)Which of the following is true about the distribution of mean times for n 38? O The distribution will be normally distributed with a mean of 59 minutes and a standard deviation of 6 minutes. O The distribution will be positively skewed with a mean of 59 minutes and a standard deviation of 0.97 minutes. The distribution will be normally distributed with a mean of 59 minutes and a standard deviation of 0.97 minutes O The distribution will be positively skewed with a mean of 59 minutes and a standard deviation of 6 minutes. (b) Calculate the probability the mean time for the sample of 38 buses will be between 58 minutes and 60 minutes. P(58 X 60) = (c) How likely is it the average time will exceeds 60 minutes? P(X 2 60)

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Answer #1

(a)

The sampling distribution of sample mean will have mean

mu_{ar{x}}=mu=59

and standard deviation

6 G- 0.97

(b)

The z-score for ar{x}=58 is

i-μί 58-59 -=-1.03 =-1.03 = 0.97

The z-score for ar{x}=60 is

- 60 59 0.97

The required probability is

P(58leq ar{x}leq 60)=P(-1.03leq zleq 1.03)=0.6970

(c)

P(F > 60)-P(z > 1.03)-1-P(s < 1.03) 0.1515

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