Question

What is the probability that the sample mean would differ from the true mean by greater than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

Answer 1

The correct solution is "0.0226".

Explanation:

The given question seems to be incomplete. Please find below the attachment of the complete query.

According to the question,

Mean

= 29

Standard deviation (s),

= 8

For sample size pf 92,

The standard error will be:

`SE=(s)/(√(N) )`

`=(8)/(√(92) )`

`=0.834`

now,

⇒
`1-P((-1.9)/(0.834) < z < (1.9)/(0.834) )` =
`1-P(-2.28<z<2.28)`

or,

=
`1-(2* P(z<2.28)-1)`

=
`2-2* P(z<2.28)`

With the help of table, the normal distribution will be:

=
`2-2* 0.9887`

=
`0.0226`