Question

Reattempt this question Question with last attempt is displayed for your review only A particular fruit's weights are normally distributed, with a mean of 327 grams and a standard deviation of 20 grams. If you pick 15 fruit at random, what is the probability that their mean weight will be between 341 grams and 343 grams. 0301 Round to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 2 decimal places are accepted.

Answer 1

Given that fruit weights are normally distributed with mean $\mu =327$ and standard deviation of $20$.

The mean of the $n=15$ fruits is $\mu =327$  and standard deviation of the mean is $\sigma =20/\sqrt{15}=5.164$ .

The probability,$P\left (341< \overline{X}<343\right )$ is found y converting to standard normal (Z) distribution.

$P\left (341< \overline{X}<343\right )=P\left (\frac{341-327}{5.164}< \frac{\overline{X}-\mu }{\sigma }<\frac{343-327}{5.164}\right )\\ P\left (341< \overline{X}<343\right )=P\left (2.7110883< Z<3.0983734\right )\\ P\left (341< \overline{X}<343\right )=\Phi \left (3.0983734\right )-\Phi \left (2.7110883\right )\\ {\color{Blue} P\left (341< \overline{X}<343\right )= 0.0024}$