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

The probability will be "0.0111".

Explanation:

The given values are:

Mean,

`\mu` = 79

Standard deviation,

`\sigma` = 3.5

Now,

⇒
`\sigma\bar x = (\sigma)/(\sqrt n)`

`= (3.5)/(\sqrt 16)`

`=0.875`

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

So,

=
`1 - P{((\bar x - \mu \bar x ))/( \sigma \bar x) < ((81 - 79) )/(0.875) ]`

=
`1 - P(z < 2.2857)`

=
`0.0111`