In a large local high school, 19% of freshmen have had their wisdom teeth removed and 24% of seniors have had their wisdom teeth removed. Suppose that a random sample of 60 freshmen and 50 seniors is selected. Let F = the proportion of freshmen who have had their wisdom teeth removed and S = the proportion of seniors who have had their wisdom teeth removed. What is the probability that the proportion of freshmen who have had their wisdom teeth removed is greater than the proportion of seniors?

Find the z-table here.

Answer:

0.2643 = 26.43% probability that the proportion of freshmen who have had their wisdom teeth removed is greater than the proportion of seniors.

Step-by-step 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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 [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

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.

In a large local high school, 19% of freshmen have had their wisdom teeth removed. Sample of 60 freshmen:

So, by the central limit theorem:

[tex]\mu_f = 0.19, s_f = \sqrt{\frac{0.19*0.81}{60}} = 0.0506[/tex]

24% of seniors have had their wisdom teeth removed. Sample of 50 seniors.

So, by the central limit theorem:

[tex]\mu_s = 0.24, s_s = \sqrt{\frac{0.24*0.76}{50}} = 0.0604[/tex]

What is the probability that the proportion of freshmen who have had their wisdom teeth removed is greater than the proportion of seniors?

This is the probability that the subtraction of the proportion of freshmen by the proportion of seniors is larger than 0. For this distribution, we have that:

[tex]\mu = \mu_f - \mu_s = 0.19 - 0.24 = -0.05[/tex]

[tex]s = \sqrt{s_f^2 + s_s^2} = \sqrt{0.0506^2 + 0.0604^2} = 0.0788[/tex]

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0 - (-0.05)}{0.0788}[/tex]

[tex]Z = 0.63[/tex]

[tex]Z = 0.63[/tex] has a pvalue of 0.7357

1 - 0.7357 = 0.2643

0.2643 = 26.43% probability that the proportion of freshmen who have had their wisdom teeth removed is greater than the proportion of seniors.

Answer

I got 0.263 and was correct.

Step-by-step explanation:

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