Question

A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally dist
If the random variable Z has a standard normal distribution, then PIZs -1.37) is:
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Answer #1

1)

If X is aptitude test score, then

P(X\geq 400)=P(\frac{X-\mu }{\sigma }\geq \frac{400-460}{80})=P(Z\geq -0.75)

P(Z\geq -0.75)=P(-0.75<Z< 0)+P(Z> 0)

Refer to standard normal distribution tables, we get

P(Z\geq -0.75)=0.2734+0.5=0.7734

Fraction of applicants who score above 400 and above= 0.7734 or 77.34%

2)

P(Z\leq -1.37)=P(Z\leq 0)-P(-1.37\leq Z<0)

Refer to standard normal distribution tables, we get

P(Z\leq -1.37)=0.5-0.4147=0.0853

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