Question

1. A variable is normally distributed in the population with a mean of 100 and a standard deviation of 10. A sample of 20 is randomly selected. The probability that the sample mean is between 90 and 110 is _______ the probability that the variable is between 90 and 110.

		greater than
		less than
		equal to
		not comparable with 2.the general manager of a logistic consulting group believes that 28% of the firm's orders come from new customers. A simple random sample of 120 orders was selected. Using the manager's belief, determine: 1. The standard error for the sampling distribution of proportion.  (3 decimal places) 2. The probability that the proportion of firm's orders that come from new customers is outside the range of 25% to 31%  (3 decimal places)

Answer 1

(1)

Correct option:

greater than

EXPLANATION:

In the case of sample mean:

Standard Error = / = 10/ = 4.4721.

For from 90 to mid value:
Z = (90 - 100)/4.4721 = - 2.2361

Table gives area = 0.4875

So,

P( between 90 and 110) = 0.4875 X 2 = 0.9750

In the case of variable:

For X from 90 to mid value:
Z = (90 - 100)/10 = - 1

Table gives area = 0.3413.

So,

P(X between 90 & 110) = 0.3413 X 2 = 0.6826

This proves that the probability that the sample mean is between 90 and 110 is greater than the probability that the variable is between 90 and 110.

(2)

n = 120

p = 0.28

q = 1 - p = 0.72

1. Standard error =

2. To find P(p<0.25) + P(p>0.31)

P(p<0.25):

Z = (0.25 - 0.28)/0.0410 = - 0.7317

Table gives area = 0.2673

So,

P(p<0.25) = 0.5 - 0.2673 = 0.2327

P(p>0.31):
Z = (0.31 - 0.28)/0.0410 = 0.7317

Table gives area = 0.2673

So,

P(p>0.31) = 0.5 - 0.2673 = 0.2327

So,

P(outside the range of 25% to 31%) = 0.2327 X 2 = 0.465

So,

Answer is:

0.465