Question

1. Given that the heights of 300 students are normally distributed with a mean of 68.0 inches and a Standard Deviation of 3.0 inches, determine how many students have heights...

(a) ... greater than 71 inches

(b) ... less than or equal to 65 inches

(c) ... between 65 inches and 71 inches inclusive

(d) ... between 59 inches and 62 inches inclusive

Assume the measurements are recorded to the nearest inch.

2. If the mean and standard deviation of a set of measurements which are normally distributed, what percentage of the data is ...

(a) ... within the range the mean +/- 2 standard deviations

(b) ... outside the range of the mean +/- 1 standard deviation

(c) ... greater than the mean - 2 standard deviations

(d) ... less than the mean + 3 standard deviations

Answer 1

Ans:

1)

a)

z=(71-68)/3

z=1

P(z>1)=0.1587%

Number of students have heights greater than 71 inches=300*0.16=48

b)

z=(65-68)/3=-1

P(z<=-1)=0.1587 %

Number of students have heights less than or equal to 65 inches=300*0.16=48

c)

P(-1<z<1)=P(z<1)-P(z<-1)=0.8413-0.1587=0.6826 %

Number of students have heights between 65 inches and 71 inches inclusive=0.68*300=204

d)

z(59)=(59-68)/3=-3

z(62)=(62-68)/3=-2

P(-3<z<-2)=P(z<-2)-P(z<-3)=0.0228-0.0013=0.0215 or 2.15%

Number of students have heights between n 59 inches and 62 inches inclusive=0.0215*300=6 approximately

2)

a)P(-2<z<2)=P(z<2)-P(z<-2)=0.9773-0.0228=0.9545 or 95.45%

b)P(z<-1 or z>1)=P(z<-1)+P(z>1)=2*0.1587=0.3174 or 31.74%

c)P(z>-2)=P(z<=2)=0.9773 or 97.73%

d)P(z<3)=0.9987 or 99.87%