Question

A particular fruit's weights are normally distributed, with a mean of 720 grams and a standard deviation of 38 grams. The heaviest 19% of fruits weigh more than how many grams? Give your answer to the nearest gram.

Answer 1

Answer: The heaviest 19% of fruits weigh more than 753grams.

Step-by-step explanation:

Let X = fruit's weights that are normally distributed.

Given: $\mu=720,\ \ \ \sigma=38$

To find : x such that P(X>x)=19%

i.e. P(X<x) = 81%  [100%-19%=81%]

i.e. P(X<x) = 0.81

$P(\dfrac{X-\mu}{\sigma}<\dfrac{x-720}{38})=0.81$

Since, $Z=\dfrac{X-\mu}{\sigma}$ and from z-table the z value for p-value of 0.81 (one -tailed) = 0.8779

$\dfrac{x-720}{38}=0.8779\\\\\Rightarrow\ x-720 =38\times0.8779\\\\\Rightarrow\ x-720 =33.36\\\\\Rightarrow\ x = 33.36+720=753.36\approx753$

Hence, the heaviest 19% of fruits weigh more than 753grams.