Question

1. A normally distributed population has a mean of 58 and a standard deviation of 18. Sample avergaes from samples of size 12 are collected. What would be the lower end of the centered interval that contains 90% of all possible sample averages?
   
   Round to the nearest hundredth

QUESTION 10
At a certain restaurant in Chicago, the average time it takes a person to eat a nice dinner is 49 minutes with a standard deviation of 18 minutes. These times are known to be normally distributed.

To Four decimal places, what is the probability a random diner will finish dinner in less than 58 minutes?

Answer 1

answer)

mean = 58

standard deviation = 18

z for 90% confidence interval is 1.645

first, we need to find the standard error

standard error = standard deviation / square root of n

n = sample size = 12

standard error = 18/ square root of 12

now we need to find the margin of error which is = z*standard error

1.645*(18/square root of 12)

=8.5477

now confidence interval is given by

lower end = mean - margin of error = 58 - 8.5477 = 49.4523

upper end = mean + margin of error = 58 + 8.5477 = 66.5477

10)

mean = 49

s = 18

we need to find

p(x<58)

for this we need to find z score

z = (x-mean)/standard deviation

= (58-49)/18

= 0.5

from z table p(z<0.5) = 0.6915

so, p(x<58) = 0.6915