Question

The mean weight of an adult is 6565 kilograms with a standard deviation of 1313 kilograms. If 9292 adults are randomly selected, what is the probability that the sample mean would be greater than 62.762.7 kilograms

Answer 1

Answer: 0.9554

Step-by-step explanation:

Let $\overline{X}$ be the sample mean.

Given: Mean weight$(\mu)$ of an adult is 65 kilograms with a standard deviation$(\sigma)$ of 13 kilograms.

Sample space = 92

The probability that the sample mean would be greater than 62.7 kilograms:

$P(\overline{X}>62.7)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{62.7-65}{\dfrac{13}{\sqrt{92}}})\\\\=P(Z>-1.70)\\\\=P(Z<1.70)\ \ \ \[P(Z>-z)=P(Z<z)]\\\\=0.9554$[ By p-value table]

Hence, the required probability= 0.9554