Question

The height distribution of NBA players follows a normal distribution with a mean of 79 inches and standard deviation of 3.5 inches. What would be the sampling distribution of the mean height of a random sample of 16 NBA players?

Answer 1

Answer:

The probability will be "0.0111".

Step-by-step explanation:

The given values are:

Mean,

$\mu$ = 79

Standard deviation,

$\sigma$ = 3.5

Now,

⇒ $\sigma\bar x = \frac{\sigma}{\sqrt n}$

         $= \frac{3.5}{\sqrt 16}$

         $=0.875$

⇒ $P(\bar x > 81) = 1 - P(\bar x < 81)$

So,

= $1 - P{\frac{(\bar x - \mu \bar x )}{ \sigma \bar x} < \frac{(81 - 79) }{0.875} ]$

= $1 - P(z < 2.2857)$

= $0.0111$