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

Answer:

weight will be greater by 20 gm.

Step-by-step 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

$z=\dfrac{x-\mu}{\sigma} \\1.18=\dfrac{x-446}{17}\\x = 466.06$

hence the weight will be greatest by

               = 466 - 446 = 20 gm

Answer 2

Answer:

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

Step-by-step 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

$Z=\frac{X-\overline{X}}{\sigma }\\\\X=\sigma \times Z+\overline{X}\\\\\therefore X=427.635grams$