# Normal distribution with the sample mean question

The diameter of an apple has mean $$8$$ cm and standard deviation $$1$$ cm. A sample of $$n$$ apples is chosen and their mean diameter measured.

What is the smallest value of $$n$$ that must be chosen if the probability of the mean diameter being between $$7.9$$ cm and $$8.1$$ cm must be at least $$0.3$$?

Here is what I've done, to no success.

$$P\left(7.9<\overline{X}<8.1\right)\ge 0.3$$

$$\frac{0.1}{\left(\frac{1}{\sqrt{n}}\right)}=0.1\sqrt{n}=z$$

$$P\left(Z\le z\right)=0.15$$ ...etc. The rest of my working doesn't lead me to the right answer too.

The sample mean $$\bar{x} \sim \mathcal{N}(8, \frac{1}{n})$$ We need $$$$P\left(7.9<\overline{X}<8.1\right)\ge 0.3$$$$ Let's standardize, $$$$P\left(\frac{7.9-8}{\frac{1}{\sqrt{n}}} Hence $$$$P( -0.1\sqrt{n} < z < 0.1\sqrt{n} ) \geq 0.3$$$$ Let's find the smallest symmetric interval $$[-\beta,\beta]$$ such that the area is less than $$0.3$$. At the limit $$$$P( -\beta < z < \beta ) = 0.3$$$$ The following is done if you only have access to a z-table giving you probabilities $$P(z < -z_0)$$ $$$$P( z < -\beta) = \frac{1-P( -\beta < z < \beta )}{2} = 0.35$$$$ Referring to the z-table, we get that $$$$\beta = 0.1 \sqrt{n} = 1.81$$$$ so $$$$\sqrt{n} = \frac{1.81}{0.1} = 18.1$$$$ So $$n = 328$$.
• I can kind of see the logic behind this how. But wouldn't it be $\frac{1}{\sqrt{n}}$ ?