Question

In​ basketball, the top free throw shooters usually have a probability of about

0.850

of making any given free throw. Over the course of a​ season, one such player shoots

300

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?

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

makes.

Answer 1

Probability, p = 0.850

n = 300

a) Considering this as binomial distribution,

Mean = n*p = 300*0.850 = 255

Standard deviation =

=

=

= 6.18466

b) According to normal distribution, 99.7% of all data fall within 3 standard deviations from the mean.

Lower limit = Mean - 3SD = 255 - 3*6.18466 236.45

Upper limit = Mean + 3SD = 255 + 3*6.18466 273.55

So, the number of free throws would almost certainly fall in the range of 236 to 274 (Rounding).

c) Proportion of free throws made at lower limit = 236.45/300 = 0.7882

Proportion of free throws made at upper limit = 273.55/300 = 0.9118

The proportion of free throws made will most certainly be between 78.82% to 91.18%.

Please comment in case of any clarifications needed. Thanks.