Question

Please help me!

Answer 1

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