Question

A particular fruit's weights are normally distributed, with a mean of 446 grams and a standard deviation of 17 grams. If a fruit is picked at random then 14% of the time, its weight will be greater than how many grams?

Answer 1

The weight corresponding to which weight will be larger than 14% of times equals 427.635 grams.

Explanation:

We need to find the value of weight that corresponds to 14% of area under the normal distribution curve

It is given that

`\overline{X}=446\\\\\sigma =17`

Using standard normal distribution tables we find value of Z corresponding to 14% of the area as -1.080

Thus using the standard equation

`Z=\frac{X-\overline{X}}{\sigma }\\\\X=\sigma * Z+\overline{X}\\\\\therefore X=427.635grams`

Answer 2

weight will be greater by 20 gm.

Explanation:

to calculate the z score use negative z table for 0.14

P ( Z < x ) = 0.86

Value of z to the cumulative probability of 0.86 from normal table is 1.18

mean(μ) = 446 gm

standard deviation(σ) = 17 grams

`z=(x-\mu)/(\sigma) \\1.18=(x-446)/(17)\\x = 466.06`

hence the weight will be greatest by

= 466 - 446 = 20 gm