Question

A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heaviest 5% of fruits weigh more than how many grams?

Answer 1

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = weights of the fruits

The z-score probability distribution for the normal distribution is given by;

                              Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = population mean weight = 733 grams

            $\sigma$ = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( $\frac{X-\mu}{\sigma}$ > $\frac{x-733}{9}$ ) = 0.05

                    P(Z > $\frac{x-733}{9}$ ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               $\frac{x-733}{9}=1.645$

                              ${x-733}}=1.645\times 9$

                              $x = 733 + 14.81$

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.