Question

A particular fruit's weights are normally distributed, with a mean of 230 grams and a standard deviation of 37 grams. If you pick 81 fruit at random, what is the probability that their mean weight will be between 231 grams and 240 grams

Answer 1

Answer: 0.3964

Step-by-step explanation:

Given: A particular fruit's weights are normally distributed, with a mean$(\mu)$ of 230 grams and a standard deviation$(\sigma)$ of 37 grams.

Sample size : n= 87

Let $\overline{X}$ be the sample mean.

The probability that their mean weight will be between 231 grams and 240 grams will be :

$P(231<\overline{X}<240)=P(\dfrac{231-230}{\dfrac{37}{\sqrt{81}}}<\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}},\dfrac{231-230}{\dfrac{37}{\sqrt{81}}})$

$=P(0.2432<Z<2.4324)\ \ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(Z<2.4324)-P(Z<0.2432)\\\\=0.9925- 0.5961=0.3964$

Hence, the required probability = 0.3964