Question

The population of weight of a particular fruit is normally distributed with a mean of 207 g and a standard deviation of 27 grams. If 16 fruits are picked at random, then 6% of the time the mean weight will be greater than how many grams

Answer 1

We have a random variable, the weight of a fruit, with a population mean of 207 g and standard deviation of 27 g.

We take a sample of 16 fruits.

We have to find the value for which the sample mean is expected to be greaterg than only 6% of the time.

To do that we start by describing the sample mean distribution.

The mean is equal to the population mean but the standard deviation is affected by the sample size:

`\begin{gathered} \mu_s=\mu=207 \\ \sigma_s=\frac{\sigma}{\sqrt[]{n}}=\frac{27}{\sqrt[]{16}}=(27)/(4)=6.75 \end{gathered}`

We now find the z-score for the standarized normal distribution for which the values are above 6%:

The value of z for this threshold value is z = 1.555.

Then, we can transform this z-score into the sample mean distribution as:

`\bar{x}=\mu_s+z\cdot\sigma_s=207+1.555\cdot27\approx249`

Answer: the sample mean will be greater than 249 grams 6% of the time.