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.