Question

Scores for a common standardized college aptitude test are normally distributed with a mean of 484 and a standard deviation of 100. Randomly selected men are given a Prepartion Course before taking this test. Assume, for sake of argument, that the Preparation Course has no effect on people's test scores.

If 1 of the men is randomly selected, find the probability that his score is at least 520.1.
P(X > 520.1) =
Enter your answer as a number accurate to 4 decimal places.

If 15 of the men are randomly selected, find the probability that their mean score is at least 520.1.
P(x-bar > 520.1) =
Enter your answer as a number accurate to 4 decimal places.

If the random sample of 15 men does result in a mean score of 520.1, is there strong evidence to support a claim that the Preapartion Course is actually effective? (Use the criteria that "unusual" events have a probability of less than 5%.)

• No. The probability indicates that is is possible by chance alone to randomly select a group of students with a mean as high as 520.1 if the Preparation Course has no effect.
• Yes. The probability indicates that is is highly unlikely that by chance, a randomly selected group of students would get a mean as high as 520.1 if the Preparation Course has no effect.

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	484
std deviation   =σ=	100.0000

probability =P(X>520.1)=P(Z>(520.1-484)/100)=P(Z>0.36)=1-P(Z<0.36)=1-0.6406=0.3594

b)

sample size       =n=	15
std error=σx̅=σ/√n=	25.8199

probability =P(X>520.1)=P(Z>(520.1-484)/25.82)=P(Z>1.4)=1-P(Z<1.4)=1-0.9192=0.0808

c)since probability is not less than 0.05:

No. The probability indicates that is is possible by chance alone to randomly select a group of students with a mean as high as 520.1 if the Preparation Course has no effect.