Answer:
0.016062.
Step-by-step explanation:
Assume that the population is normal:
~(,
2
) = (1185, 702
).
Then the distribution of the sample mean
̅ =
1 + 2 + 3 + ⋯ + 100
100
is exactly normal with mean
̅= (̅) = = 1185 hours
and standard deviation
̅= (̅) =
√
=
70
√100
= 7 hours.
The standardized variable
=
̅ − ̅
̅
=
̅ − 1185
7
Is distributed as (0,1).
The following value of corresponds to the value ̅= 1200 of ̅:
=
̅−̅
̅
=
1200−1185
7
= 2.142857.
Therefore,
(̅ ≥ 1200) = (
̅ − ̅
̅
≥
1200 − ̅
̅
) = ( ≥
1200 − 1185
7
) = ( ≥ 2.142857) =
= 1 − ( < 2.142857) = 1 − 0.983938 = 0.016062,
because using the command
= NORM. S.DIST(2,142857; TRUE)
from Microsoft Excel we can see that
= 2.142857
gives
( < 2.142857) = 0.983938.
Only rarely, just over one time in a hundred tries of 100 light bulbs, would the average life exceed 1200 hours
