Question

The population of weights of a particular fruit is normally distributed, with a mean of 464 grams and a standard deviation of 6 grams. If 27 fruits are picked at random, then 2% of the time, their mean weight will be greater than how many grams?

Answer 1

Answer:

If 27 fruits are picked at random, then 2% of the time, their mean weight will be greater than 466.37 grams

Step-by-step explanation:

Mean = $\mu = 464$

Standard deviation =$\sigma = 6$

We are supposed to find If 27 fruits are picked at random, then 2% of the time, their mean weight will be greater than how many grams i.e.P(X>x)=0.02

The mean weight is in the highest 2%, you want to go to a z-table and find the z-score that where the area to the left of the curve is closest to 0.98.

n = 27

Refer the z -table

P(Z>x)=2.06

$\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}=2.06\\\frac{x-464}{\frac{6}{\sqrt{27}}}=2.06\\x-464=2.06 \times \frac{6}{\sqrt{27}}\\x=(2.06 \times \frac{6}{\sqrt{27}})+464\\x=466.37$

So, If 27 fruits are picked at random, then 2% of the time, their mean weight will be greater than 466.37 grams