Question

A particular fruit's weights are normally distributed, with a mean of 745 grams and a standard deviation of 21 grams.

The heaviest 9% of fruits weigh more than how many grams?

Give your answer to the nearest gram.

Check Answer Question 9 A particular fruit's weights are normally distributed, with a mean of 745 grams and a standard deviation of 21 grams. The heaviest 9% of fruits weigh more than how many grams? Give your answer to the nearest gram. Calculator Check Answer

Answer 1

Solution:-

Given that,

mean = $\mu$ = 745

standard deviation = $\sigma$ = 21

Using standard normal table,

P(Z > z) = 9%

= 1 - P(Z < z) = 0.09  

= P(Z < z) = 1 - 0.09

= P(Z < 1.341 ) = 0.91  

z = 1.341

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 1.341 * 21 + 745

x = 773.16

x = 773 grams.

Answer 2

SOLUTION :

Let a grams be the weight  over which 9% fruits are there.

P( x > a) = P( z > (a - m)/s)

=> 0.09 =  P( z > (a - m)/s)

From ND table : 

z > 1.34 

The cutoff point is z = 1.34

So, 

a = z * s + m = 1.34 * 21 + 745 = 773.14 = 773 grams

So, the heaviest 9% fruits weigh more than 773 grams (ANSWER).