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?

Respuesta :

Answer:

The probability will be "0.0111".

Step-by-step explanation:

The given values are:

Mean,

[tex]\mu[/tex] = 79

Standard deviation,

[tex]\sigma[/tex] = 3.5

Now,

⇒ [tex]\sigma\bar x = \frac{\sigma}{\sqrt n}[/tex]

         [tex]= \frac{3.5}{\sqrt 16}[/tex]

         [tex]=0.875[/tex]

⇒ [tex]P(\bar x > 81) = 1 - P(\bar x < 81)[/tex]

So,

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

= [tex]1 - P(z < 2.2857)[/tex]

= [tex]0.0111[/tex]