The answers are a) 15.71% b) 34.46% c) 99.5 and d) (105.5,114.5). Please explain

Question

The answers are a) 15.71% b) 34.46% c) 99.5 and d) (105.5,114.5). Please explain how to solve the problem step by step.

In a college, students' scores of their entrance exams are normally distributed with a mean of 110 (out of 160) and a standard deviation of 12.5 (a) What percent of students have their entrance exam scores between 90 and 100? (b) What percent of students have their entrance exam scores below 105? (c) What is the cutoff for the lowest 20% exam scores? (d) What is the range for the middle 28% exam scores?

Explanation / Answer

a) P((90-110)/12.5<z<(100-110)/12.5)

= P(-1.6<z<-0.8)

Use Excel function

NORMSDIST(-0.8)-NORMSDIST(-1.6)

= 0.1571 OR 15.71%

B) P(Z<(105-110)/12.5)

= P(Z<-0.4)

Use Excel function

NORMSDIST(-0.4)

= 0.3446 or 34.46%

c) Lowest 20% = NORMSINV(0.20)*12.5+110 = 99.5

d) Lower value = NORMSINV(0.36)*12.5+110 = 105.5

Upper value = NORMSINV(0.64)*12.5+110 = 114.5

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