Answer:
Answer and Explanation:
We have:
Population mean,
μ
=
3
,
000
hours
Population standard deviation,
σ
=
696
hours
Sample size,
n
=
36
1) The standard deviation of the sampling distribution:
σ
¯
x
=
σ
√
n
=
696
√
36
=
116
2) As per the central limit theorem, the expected value of the sampling distribution is equal to the population mean.
Therefore:
The expected value of the sampling distribution is equal to the population mean,
μ
¯
x
=
μ
=
3
,
000
The standard deviation of the sampling distribution,
σ
¯
x
=
116
The shape of the sampling distribution of
¯
x
is approximately normal. As the sample size is more than
30
.
3) The probability that the average life in the sample will be between
2670.56
and
2809.76
hours:
P
(
2670.56
<
x
<
2809.76
)
=
P
(
2670.56
−
3000
116
<
z
<
2809.76
−
3000
116
)
=
P
(
−
2.84
<
z
<
−
1.64
)
=
P
(
z
<
−
1.64
)
−
P
(
z
<
−
2.84
)
=
0.0482
Using Excel: =NORMSDIST(-1.64)-NORMSDIST(-2.84)
4) The probability that the average life in the sample will be greater than
3219.24
hours:
P
(
x
>
3219.24
)
=
P
(
z
>
3219.24
−
3000
116
)
=
P
(
z
>
1.89
)
=
0.0294
Using Excel: =NORMSDIST(-1.89)
5) The probability that the average life in the sample will be less than
3180.96
hours:
P
(
x
<
3180.96
)
=
P
(
z
<
3180.96
−
3000
116
)
=
P
(
z
<
1.56
)
=
0.9406
