Question

Suppose that X is a normal random variable, with its mean of 3 and its standard deviation of 2. (a) Suppose that David is going to randomly choose a sample of 10 observations from this population. What is the probability that the sample mean of X exceeds 3.2

Answer 1

Answer:   0.376

Step-by-step explanation:

Given: X is a normal random variable, with its mean$(\mu)$ of 3 and its standard deviation$(\sigma)$ of 2.

Sample size : n= 10

The probability that the sample mean of X exceeds 3.2 will be :

$P(\overline{X}>3.2)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{3.2-3}{\dfrac{2}{\sqrt{10}}})\\\\=P(Z>0.316)\\\\=1-P(Z<0.316)\\\\=1-0.6240 =0.376$

Hence, the probability that the sample mean of X exceeds 3.2= 0.376