Answer:
0.284
Explanation:
To carry out this calculation, we begin by describing the sampling distribution of the sample proportion.
The sample size is n = 50 and the population proportion of teachers who made an apparel purchase is 0.56.
Shape: Because np = (50)(0.56) = 28 and n(1 – p) = (50)(0.44) = 22 are both at least 10, the shape of the sampling distribution of the sample proportion is approximately Normal.
Center:
μ
p
^
=
p
=
0
.
5
6
μ
p
^
=p=0.56
Variability: The standard deviation of the sample proportion is approximately
(
0
.
5
6
)
(
0
.
4
4
)
5
0
≈
0
.
0
7
0
2
50
(0.56)(0.44)
≈0.0702.
P(
p
^
p
^
> 0.6) = Normalcdf(lower: 0.6, upper: 1000, mean: 0.56, SD: 0.0702) = 0.284.
P
(
p
^
>
0
.
6
)
=
P
(
z
>
0
.
6
−
0
.
5
6
0
.
0
7
0
2
)
=
P
(
z
>
0
.
5
7
)
=
1
−
0
.
7
1
5
7
=
0
.
2
8
4
3
P(
p
^
>0.6)=P(z>
0.0702
0.6−0.56
)=P(z>0.57)=1−0.7157=0.2843