Question

Suppose a random sample of n = 16 observations is selected from a population that is normally distributed with mean equal to 101 and standard deviation equal to 12.
(a) Give the mean and the standard deviation of the sampling distribution of the sample mean x(with a line above). (Enter your standard deviation to two decimal places.)

mean =

standard deviation =

(b) Find the probability that x(line above it) exceeds 107. (Round your answer to four decimal places.)

Answer:

(c) Find the probability that the sample mean deviates from the population mean μ = 101 by no more than 4. (Round your answer to four decimal places.)

Answer 1

Answer:

101, 3, 0.025, 0.7416

Step-by-step explanation:

Given that  a random sample of n = 16 observations is selected from a population that is normally distributed with mean equal to 101 and standard deviation equal to 12.

As per central limit theorem we have

a) Mean of sample mean = $E(\bar x) =\\ \mu =101$

Std deviation of sample mean = $\frac{\sigma}{\sqrt{n} } =3$

Mean = 101

Std dev =3.00

b) $P(\bar x >107) = P(Z>\frac{107-101}{3} )\\= P(Z>2) = 0.025$

c) the probability that the sample mean deviates from the population mean μ = 101 by no more than 4.

=$P(|\bar x-101|) \leq 4\\= P(|z|\leq 1.13)\\= 2(0.3708)\\=0.7416$