Mathematics, 30.03.2020 21:13 02s54mcoupe

In college basketball, a shot made from beyond a designated arc radiating about 20 ft from the basket is worth three points instead of the usual two points for shots made inside that arc. Over his career, University of Florida player Lee Humphrey made 45% of his three-point attempts. In his final season, he made 25 of 100 three-point shots, leading to a tv basketball analyst to announce that Humphrey was in a shooting slump. a) Assuming Humphrey has a 45% chance of making any particular three-point shot, find the mean and standard deviation of the sampling distribution of the proportion of three point shots he will make (3 points) b) How many standard deviations from the mean is his result of making 25 of 100 shots? (3 points) c) If Humphrey was actually not in a slump but still had a 45% chance of making any particular three-point shot, explain if it would or would not be surprising for him to only make 25 of 100 shots. Why? (3 points)

Answers: 3

Show answers

Answers

a)By the Central Limit Theorem, the distribution would be approximately normal, with mean $\mu = 0.45$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497$

b)4.02 standard deviations below the mean.

c)Making 25 or less shots has a pvalue of -4.02. Z-scores lower than -2 are considered surprising, so yes, it would be surprising for him to make only 25 of these shots.

Step-by-step explanation:

To solve this question, we are going to need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For proportions, we have that the mean is $\mu = p$ and the standard deviation is $s = \sqrt{\frac{p(1-p)}{n}}$

In this problem, we have that:

p = 0.45, n = 100

a)By the Central Limit Theorem, the distribution would be approximately normal, with mean $\mu = 0.45$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497$

b)This is Z when X = 0.25.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$