Question

Suppose a random sample of n = 25 observations is selected from a population that is normally distributed with mean equal to 108 and standard deviation equal to 14.

(a) Give the mean and the standard deviation of the sampling distribution of the sample mean

x.

mean
standard deviation

(b) Find the probability that

x

exceeds 113. (Round your answer to four decimal places.)


(c) Find the probability that the sample mean deviates from the population mean ? = 108 by no more than 2. (Round your answer to four decimal places.)

Answer 1

(a)

The mean and the standard deviation of the sampling distribution of the sample mean from the normal population with mean and standard deviation are and respectively. Here , and n = 25. So the mean of the sampling distribution of the sample mean X is 108 and the standard deviation is

X

Mean = 108

Standard deviation = 2.8

(b)

The probability that X exceeds 113 = P(X > 113). Using the z-transformation,

The probability that X exceeds 113 is 0.0367

(c)

The probability that the sample mean deviates from the population mean 108 by no more than 2 can be denoted as P(107 < X < 109).

= P(-0.36 < Z <0.36)

= 2P(Z<0.36) - 1

= 2(0.64058)-1

=0.2812

The probability that the sample mean deviates from the population mean 108 by no more than 2 is 0.2812