Question

A study involves a population of 400 tall women. This population has a mean height of 179.832 cm and a standard deviation of 12.192 cm. If 50 of these women are randomly selected, find the probability that the mean for this sample group is above 182.88

Answer 1

Answer:  0.0386

Step-by-step explanation:

Given: The population of 400 tall women has a mean height$(\mu)$ of 179.832 cm and a standard deviation$(\sigma)$ of 12.192 cm.

Let X be a random variable that represents the height of woman.

Sample size : n= 50

The probability that the mean for this sample group is above 182.88 will be :

$P(\overline{X}>182.88)\\\\=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{182.88-179.832}{\dfrac{12.192}{\sqrt{50}}})\\\\ =P(Z>1.7678)\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-P(Z<1.7678)\\\\=1-0.9614\ \ \ [\text{By p-value table}]\\\\= 0.0386$

Hence, Required probability =  0.0386