Question

One measure of an athlete’s ability is the height of his or her vertical leap. Many professional basketball players are known for their remarkable vertical leaps, which lead to amazing dunks. DJ Stephens is the current vertical-leap record holder at 46 inches. However, the mean vertical leap of all NBA players is 28 inches. Suppose the standard deviation is 7 inches and 36 NBA players are selected at random.

1. The probability that the mean vertical leap for the 36 players will be less than 26 inches is ________

2. The probability that the mean vertical leap for the 26 players will be between 27.5 and 28.5 inches is _______

Answer 1

Answer:

(1) P($\bar X$ < 26 inches) = 0.0436

(2) P(27.5 inches < $\bar X$ < 28.5 inches) = 0.2812

Step-by-step explanation:

We are given that the mean vertical leap of all NBA players is 28 inches. Suppose the standard deviation is 7 inches and 36 NBA players are selected at random.

Firstly, Let $\bar X$ = mean vertical leap for the 36 players

Assuming the data follows normal distribution; so the z score probability distribution for sample mean is given by;

            Z = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean vertical  leap = 28 inches

            $\sigma$ = standard deviation = 7 inches

            n = sample of NBA player = 36

(1) Probability that the mean vertical leap for the 36 players will be less than 26 inches is given by = P($\bar X$ < 26 inches)

   P($\bar X$ < 26) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{26-28}{\frac{7}{\sqrt{36} } }$ ) = P(Z < -1.71) = 1 - P(Z $\leq$ 1.71)

                                                 = 1 - 0.95637 = 0.0436

(2) Now, here sample of NBA players is 26 so n = 26.

Probability that the mean vertical leap for the 26 players will be between 27.5 and 28.5 inches is given by = P(27.5 inches < $\bar X$ < 28.5 inches) = P($\bar X$ < 28.5 inches) - P($\bar X$ $\leq$ 27.5 inches)

    P($\bar X$ < 28.5) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{28.5-28}{\frac{7}{\sqrt{26} } }$ ) = P(Z < 0.36) = 0.64058 {using z table}                      

    P($\bar X$ $\leq$ 27.5) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{27.5-28}{\frac{7}{\sqrt{26} } }$ ) = P(Z $\leq$ -0.36) = 1 - P(Z < 0.36)

                                                        = 1 - 0.64058 = 0.35942

Therefore, P(27.5 inches < $\bar X$ < 28.5 inches) = 0.64058 - 0.35942 = 0.2812