Question

The number of nails of a given length is normally distributed with a mean length of 5 in. and a standard deviation of 0.03 in. In a bag containing 120 nails, how many nails are more than 5.03 in. long?

a.about 38 nails
b.about 41 nails
c.about 16 nails
d.about 19 nails

Answer 1

(D) About 19 nails

Explanation:

It is given that mean length= 5 inches and the standard deviation =0.03.

In order to standardize x to z, we use the formula=
`\frac{(x-{\mu})}{{\sigma}}`.

Now, probability of nails having length more than 5.03=
`P(x>5.03)=P(z>((5.03-5))/(0.03))`

=
`P(z>1)`

=
`0.1587` b(Using the normal probability table)

In a bag of 20 nails, nails having length more than 5.03 inches=
`20{*}0.1587`=
`19.004`

= about
`19` nails.

Answer 2

The length 5.03 inches belonged to the first range of span of standard deviation. The probability of having length below 5.03 is equal to 68%. The proprobability of having length above 5.03 therefore is (100-68)/2 or equal to 16%. 16% of 120 nails given is 19.2 or D. about 19 nails.