Question

Ricardo and Tammy practice putting golf balls. Ricardo makes 47% of his putts and Tammy makes 51% of her putts. Suppose that Ricardo attempts 25 putts and Tammy attempts 30 putts. Let R = the proportion of putts Ricardo makes and T = the proportion of putts Tammy makes. What is the probability that Ricardo makes a higher proportion of putts than Tammy?

Find the z-table here.

Answer 1

A (.384)

Explanation:

EDGE2021! :)

Answer 2

0.3821 = 38.21% probability that Ricardo makes a higher proportion of putts than Tammy.

Explanation:

To solve this question, we need to understand the normal distribution, the central limit theorem, and subtraction of normal variables.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

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

`Z = (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 = (\sigma)/(√(n))`.

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

Subtraction of normal variables:

When we subtract normal variables, the mean of the subtraction will be the subtraction of the means, while the standard deviation will be the square root of the sum of the variances.

Ricardo makes 47% of his putts, and attempts 25 putts.

By the Central Limit Theorem, we have that:

`\mu_R = 0.47, s_R = \sqrt{(0.47*0.53)/(25)} = 0.0998`

Tammy makes 51% of her putts, and attempts 30 putts.

By the Central Limit Theorem, we have that:

`\mu_T = 0.51, s_T = \sqrt{(0.51*0.49)/(30)} = 0.0913`

What is the probability that Ricardo makes a higher proportion of putts than Tammy?

This is the probability that the subtraction of R by T is larger than 0. The mean and standard deviation of this distribution are, respectively:

`\mu = \mu_R - \mu_T = 0.47 - 0.51 = -0.04`

`s = √(s_R^2 + s_T^2) = √(0.0998^2 + 0.0913^2) = 0.1353`

This probability is 1 subtracted by the pvalue of Z when X = 0. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (0 - (-0.04))/(0.1353)`

`Z = 0.3`

`Z = 0.3` has a pvalue of 0.6179

1 - 0.6179 = 0.3821

0.3821 = 38.21% probability that Ricardo makes a higher proportion of putts than Tammy.