1)a)
| 
for normal distribution z score =(X-μ)/σ | 
| 
 | 
 | 
| 
here mean=       μ= | 
43.7 | 
| 
std deviation   =σ= | 
4.2000 | 
| 
probability = | 
P(X<40) | 
= | 
P(Z<-0.88)= | 
0.1894 | 
b)
| 
sample size       =n= | 
9 | 
| 
std error=σx̅=σ/√n= | 
1.4000 | 
| 
probability = | 
P(X<40) | 
= | 
P(Z<-2.64)= | 
0.0041 | 
c)
| 
sample size       =n= | 
15 | 
| 
std error=σx̅=σ/√n= | 
1.0844 | 
| 
probability = | 
P(X>46) | 
= | 
P(Z>2.12)= | 
1-P(Z<2.12)= | 
1-0.9830= | 
0.0170 | 
2)
| 
for normal distribution z score =(X-μ)/σ | 
| 
 | 
 | 
| 
here mean=       μ= | 
90 | 
| 
std deviation   =σ= | 
10.0000 | 
| 
probability = | 
P(X>95) | 
= | 
P(Z>0.5)= | 
1-P(Z<0.5)= | 
1-0.6915= | 
0.3085 | 
b)
| 
sample size       =n= | 
24 | 
| 
std error=σx̅=σ/√n= | 
2.0412 | 
| 
probability = | 
P(X>95) | 
= | 
P(Z>2.45)= | 
1-P(Z<2.45)= | 
1-0.9929= | 
0.0071 | 
c)
| 
sample size       =n= | 
20 | 
| 
std error=σx̅=σ/√n= | 
2.2361 | 
| 
probability = | 
P(X>93) | 
= | 
P(Z>1.34)= | 
1-P(Z<1.34)= | 
1-0.9099= | 
0.0901 | 
as probability of that happening is not less than 0.05 level ;
therefore it is not unusual.