Question

A basketball coach is looking over the possessions per game during last season. Assume that the possessions per game follows an unknown distribution with a mean of 56 points and a standard deviation of 12 points. The basketball coach believes it is unusual to score less than 50 points per game. To test this, she randomly selects 36 games. Use a calculator to find the probability that the sample mean is less than 50 points. Round your answer to three decimal places if necessary.

Answer 1

Answer:

The probability that the sample mean is less than 50 points = 0.002    

Step-by-step explanation:

Step(i):-

Given mean of the normal distribution = 56 points

Given standard deviation of the normal distribution = 12 points

Random sample size 'n' = 36 games

Step(ii):-

Let x⁻ be the random variable of normal distribution

Let x⁻ = 50

$Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{50-56 }{\frac{12}{\sqrt{36} } }= -3$

The probability that the sample mean is less than 50 points

P( x⁻≤ 50) = P( Z≤-3)

                = 0.5 - P(-3 <z<0)

               = 0.5 -P(0<z<3)

               =  0.5 - 0.498

               = 0.002

Final answer:-

The probability that the sample mean is less than 50 points = 0.002