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

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(\frac{\overline{X}-\mu}{(\sigma)/(√(n))}>(3.2-3)/((2)/(√(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