Question

Find the probability in each case. Appendix A Statistical Tables (Round the values of z to 2 decimal places. Round your answers to 4 decimal places.) a. ? N = 2,000 , n = 40 , ? = 74 , and , ? ? = 6 ; P ( x ¯ < 75.4 ) = ? b. ? N = 70 , n = 36 , ? = 105 , and , ? ? = 3.45 ; P ( 104 < x ¯ < 104.3 ) = ? c. ? N = 252 , n = 104 , ? = 36.0 , and , ? ? = 4.87 ; P ( x ¯ ? 37 ) = ? d. ? N = 6,000 , n = 30 , ? = 126 , and , ? ? = 13.7 ; P ( x ¯ ? 122 ) = ?

Answer 1

a)
mean= 74
sd= 6
n= 40

P(X < 75.4)
I know that, z = (X-mean)/(sd/sqrt(n))
z1 = (75.4-74)/6/sqrt(40))
z1= 1.4800

hence,
P(X < 75.4)=
= P(Z<1.48)
NORMSDIST(1.48)
0.9306

b)
mean= 105
sd= 3.45
n= 36

P(104 < X < 104.3)
= P(X<104.3) - P(X<104)

I know that, z = (X-mean)/(sd/sqrt(n))
z1 = (104-105)/3.45/sqrt(36))
z1= -1.7400
z2 = (104.3-105)/3.45/sqrt(36))
z2= -1.2200

hence,
P(104 < X < 104.3)=
= P(Z<-1.22) - P(Z<-1.74)
= NORMSDIST(-1.22) - NORMSDIST(-1.74)
0.0703

c)
mean= 36
sd= 4.87
n= 104
P(X>37)
= 1 - P(X<37)

I know that, z = (X-mean)/(sd/sqrt(n))
z1 = (37-36)/4.87/sqrt(104))
z1= 2.09

hence,
P(X>37)=
= 1 - P(Z<2.09)
= 1 - NORMSDIST(2.09)
0.0183

d)
mean= 126
sd= 13.7
n= 30

P(X < 122)
I know that, z = (X-mean)/(sd/sqrt(n))
z1 = (122-126)/13.7/sqrt(30))
z1= -1.5992

hence,
P(X < 122)=
= P(Z<-1.5992)
NORMSDIST(-1.5992)
0.0549