Question

Suppose a random sample of n = 25 observations is selected from a population that is normally distributed with mean equal to 106 and standard deviation equal to 15.

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

mean=
standard deviation=

(b) Find the probability that x̄ exceeds 115. (Round your answer to four decimal places.)


(c) Find the probability that the sample mean deviates from the population mean μ = 106 by no more than 5. (Round your answer to four decimal places.)

Answer 1

Suppose a random sample of n = 25 observations is selected from a population that is normally distributed with mean equal to 106 and standard deviation equal to 15.

(a) Mean of the sample, x̄ = mean of the population = 106
Standard deviation of the sampling distribution,
$\sigma _{x}=\frac{\sigma }{\sqrt{n}}=\frac{15}{\sqrt{25}}=3$
(b)The probability that x̄ exceeds 115

Using, 2= T- o/

$P(\bar{x}>115)=P(z>\frac{115-106}{3})=P(Z>3)$

P(x̄ > 115) = 0.0013. (Using z table)

(c) Probability that the sample mean deviates from the population mean μ = 106 by no more than 5
= P(106 - 5 < x̄ < 106 + 5) = P(101 < x̄ < 111)

101 = PG - 106 - 106 3 P(-1.6666<Z< 1.6666) 3

= 0.9044