Question

Find the value of the interval of X vbetween 80 to 90 points for the exam scores with normal distribution defined by mean of 116, and standard deviation of 16.

Find the value of x (score) for given percent of top 10% of applicants to be admitted in the police academy, where scores are normally distributed with the mean of 120 and standard deviation of 20.

Answer 1

1) $\mu=116,\ \sigma=16$

$P(80<X<90)=P(\frac{80-\mu}{\sigma}<Z<\frac{90-\mu}{\sigma})=P(\frac{80-116}{16}<Z<\frac{90-116}{16})=$

$=P(-2.25<Z<-1.625)=P(Z<-1.625)-P(Z<-2.25)=$

$=0.0526-0.0122=0.0404=4.04\%$

2) $\mu=120,\ \sigma=20$

$P(X<x)=90\%=0.9 \implies P(Z<\frac{x-\mu}{\sigma})=P(Z<\frac{x-120}{20})=0.9$

$\frac{x-120}{20}=1.28\implies x=146$

Answer:

4.04%

146