Question

A particular fruit's weights are normally distributed, with a mean of 498 grams and a standard deviation of 32 grams.If you pick 13 fruits at random, then 11% of the time, their mean weight will be greater than how many grams?Give your answer to the nearest gram.

Answer 1

We are given the following information

Mean = 498 grams

Standard deviation = 32 grams

Sample size = 13

We are asked to find the weight for which 11% of the fruits are heavier.

11% corresponds to 100% - 11% = 89% percentile of the normal distribution.

Recall that the z-score for a normal distribution is given by

`z=\frac{x-\mu}{\frac{\sigma}{\sqrt[]{n}}}`

Where x is the value that we need to find out, μ is the mean, σ is the standard deviation, n is the sample size.

z is the value of the z-score corresponding to the 89th percentile.

From the z-table, the value of the z-score corresponding to the p = 0.89 is 1.23

Let us substitute all these values into the above formula and solve for x.

`\begin{gathered} 1.23=\frac{x-498}{\frac{32}{\sqrt[]{13}}} \\ 1.23\cdot\frac{32}{\sqrt[]{13}}=x-498 \\ 1.23\cdot\frac{32}{\sqrt[]{13}}+498=x \\ 10.92+498=x \\ 509=x \\ x=509\; \text{grams} \end{gathered}`

Therefore, the required value is 509 grams.