Question

scores for a common standardized college aptitude test are normally distributed with a mean of 506 and a standard deviation of 114. randomly selected students are given a test preparation course before taking this test. assume, for sake of argument, that the preparation course has no effect. do not round between steps and use technology. round answers to 4 decimal places.
a) if 1 student is randomly selected, find the probability that their score is at least 582.2.
p(x > 582.2) =

b) if 14 students are randomly selected, find the probability that their mean score is at least 582.2.
p(\overline{x} > 582.2) =

c) assume that any probability less than 5% is sufficient evidence to conclude that the preparation course does help students perform better on the test. if the random sample of 14 students does result in a mean score of 582.2, is there strong evidence to support the claim that the course is actually effective?
○ yes. the probability indicates that it is (highly?) unlikely that by chance, a randomly selected group of students would get a mean as high as 582.2.
○ no. the probability indicates that it is possible by chance alone to randomly select a group of students with a mean as high as 582.2.

Explanation:

Step1: Calculate the z - score for part a

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Given \(\mu = 506\), \(\sigma=114\), and \(x = 582.2\).

$$z=\frac{582.2 - 506}{114}=\frac{76.2}{114}=0.6684$$

We want \(P(X>582.2)\), which is equivalent to \(P(Z > 0.6684)\). Using the property \(P(Z>z)=1 - P(Z\leq z)\). Looking up \(P(Z\leq0.6684)\) in the standard normal table or using technology (e.g., in R: pnorm(0.6684)), we get \(P(Z\leq0.6684)\approx0.7486\). So \(P(Z > 0.6684)=1 - 0.7486=0.2514\)

Step2: Calculate the z - score for part b

The formula for the z - score of the sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Given \(\mu = 506\), \(\sigma = 114\), \(n = 14\), and \(\bar{x}=582.2\)

$$z=\frac{582.2-506}{\frac{114}{\sqrt{14}}}=\frac{76.2}{\frac{114}{3.7417}}$$

$$=\frac{76.2\times3.7417}{114}=\frac{285.1175}{114}\approx2.5$$

We want \(P(\bar{X}>582.2)\), which is equivalent to \(P(Z > 2.5)\). Using the property \(P(Z>z)=1 - P(Z\leq z)\). Looking up \(P(Z\leq2.5)\) in the standard normal table or using technology (e.g., in R: pnorm(2.5)), we get \(P(Z\leq2.5)=0.9938\). So \(P(Z > 2.5)=1 - 0.9938 = 0.0062\)

Step3: Answer part c

Since the probability \(P(\bar{X}>582.2)=0.0062<0.05\)

Answer:

a) \(0.2514\)
b) \(0.0062\)
c) Yes. The probability indicates that it is (highly?) unlikely that by chance, a randomly selected group of students would get a mean as high as \(582.2\).