Question

380 free throws u In basketball, the top free throw shooters usually have a probability of about 0.90 of making any given free throw. Over the course of a season, one such player shoots 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 --(Type an integer or a decimal.) (Round to three decimal places as needed.) bli b. By the normal distribution approximation, within what range would the number of free throws made almost certainly fall, if p - 0.907 The number of free throws made would almost certainly fall between and di (Use ascending order. Round to the nearest Integer as needed.) c. Within what range would the proportion of free throws made be expected to fall? u oli The proportion of free throws made would almost certainly fall between and (Use ascending order. Round to two decimal places as needed) ert

Answer 1

Answer:

a)

Given,

sample n = 380

p = 0.90

Mean = np = 380*0.90 = 342

Standard deviation = sqrt(npq) = sqrt(380*0.90*(1-0.90)) = 5.848

b)

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

Lower limit = Mean - 3*SD = 342 - 3*5.848 = 325

Upper limit = Mean + 3*SD = 342 + 3*5.848 = 360

So, the number of free throws would almost certainly fall in the range of 325 to 360.

c)

Proportion of free throws made at lower limit = 325/380 = 0.8553

Proportion of free throws made at upper limit = 360/380 = 0.9474

The proportion of free throws made will most certainly be between 85.53% to 94.74%.