Question

2.   A variable is normally distributed with a mean of 120 and a standard deviation of 15. Twenty five scores are randomly sampled.

a)   What is the probability that the mean of the four scores is above 111?

b)   What is the probability that the mean of the four scores will be between 114 and 123?

4.Suppose that a sample of 36 employees at a large company were surveyed and asked how many hours a week they thought the company wasted on unnecessary meetings. The mean number of hours these employees stated was 12.4 with a sample standard deviation of 5.1.

Calculate a 95% confidence interval to estimate the mean amount of time all employees at this company believe is wasted on unnecessary meetings each week.

5.An automobile dealer wants to estimate the proportion of customers who still own the cars they purchased 5 years earlier. A random sample of 300 customers selected from the auto mobile dealers records indicates that 82 still own cars that were purchased 5 years earlier. Set up a 99% confidence interval estimate of the population proportion of all customers who still own the cars 5 years after they were purchased

6.If in a sample of size n=49 selected normal population, the sample mean X=48 and population standard deviation is 21, what is your statistical decision, if your H0 :µ = 45, α=0.05

3.   The mean delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is 81 restaurants with the same sample mean. Find a 90% confidence interval estimate for the population mean delivery time.

Answer 1

Solution:

2. Given that mean = 120, sd = 15, n = 25

a) P(X > 111) = P((X-mean)/(sd/sqrt(n)) > (111-120)/(15/sqrt(25)))
= P(Z > -3)
= P(Z < 3)
= 0.9987

b) P(114 < X < 123) = P((114-120)/(15/sqrt(25)) < Z < (123-120)/(15/sqrt(25)))
= P(-2 < Z < 1)
= P(Z < 1) - P(Z < -2)
= 0.8413 - 0.0228
= 0.8185

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