Question

In​ basketball, the top free throw shooters usually have a probability of about 0.80 of making any given free throw. Over the course of a​ season, one such player shoots 330 free throws.

a) Find the mean and standard deviation of the probability distribution of the number of free throws he makes.

(b) By the normal distribution​ approximation, within what range would the number of free throws made almost certainly​ fall? Why?

(c) Within what range would the proportion made be expected to​ fall?

Answer 1

We have:n=330

p=0.80

q=1-p=1-0.80=0.20

(A)

X~binom(330,0.80)

mean=np=330*0.80=264

standard deviation=$\sqrt{npq}=\sqrt{330*0.80*0.20}=\sqrt{52.8}=7.266$

(B)

X~N(np,npq)

X~N(264,7.266²)

X~N(264,52.79)

For range we calculate the confidence interval for single mean to get within what values the the number of free throws fall:

We need to construct the 95% confidence interval for the population mean \muμ. The following information is provided:

Sample Mean Xˉ =	264
Population Standard Deviation (σ) =	7.266
Sample Size(N) =	330

The critical value for α=0.05 is $z_c = z_{1-\alpha/2} = 1.96$ The corresponding confidence interval is computed as shown below:

CI​=(263.216,264.784)​

7.266 CI X - 2c X + ze + х Tin 264 – 1.96 7.266 264 +1.96 1330 (263.216, 264.784) 330

Hence the range 263.216<RANGE<264.784

(C)

For range we calculate the confidence interval for single proportion to get within what values the the number of free throws fall:

We need to construct the 95% confidence interval for the population proportion. We have been provided with the following information about the sample proportion:

Sample Proportion p^​ =	0.80
N =	330

The critical value for α=0.05 is $z_c = z_{1-\alpha/2} = 1.96$ . The corresponding confidence interval is computed as shown below:

$\begin{array}{ccl} CI(\text{Proportion}) =\displaystyle \left( \hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}} \right) =\displaystyle \left( 0.8 - 1.96 \times \sqrt{\frac{0.8 (1- 0.8)}{330}}, 0.8 + 1.96 \times \sqrt{\frac{0.8 (1- 0.8)}{330}} \right) = (0.757, 0.843) \end{array}$CI(Proportion)​=(0.757,0.843)​

Hence Range 0.757<RANGE <0.843

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