Answer: 0.376
Step-by-step explanation:
Given: X is a normal random variable, with its mean[tex](\mu)[/tex] of 3 and its standard deviation[tex](\sigma)[/tex] of 2.
Sample size : n= 10
The probability that the sample mean of X exceeds 3.2 will be :
[tex]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[/tex]
Hence, the probability that the sample mean of X exceeds 3.2= 0.376
