Normal distribution with the sample mean question

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.

Answer 1

The sample mean $$\bar{x} \sim \mathcal{N}(8, \frac{1}{n})$$ We need \begin{equation} P\left(7.9<\overline{X}<8.1\right)\ge 0.3 \end{equation} Let's standardize, \begin{equation} P\left(\frac{7.9-8}{\frac{1}{\sqrt{n}}}<z<\frac{8.1-8}{\frac{1}{\sqrt{n}}} \right)\ge 0.3 \end{equation} Hence \begin{equation} P( -0.1\sqrt{n} < z < 0.1\sqrt{n} ) \geq 0.3 \end{equation} Let's find the smallest symmetric interval $[-\beta,\beta]$ such that the area is less than $0.3$. At the limit \begin{equation} P( -\beta < z < \beta ) = 0.3 \end{equation} The following is done if you only have access to a z-table giving you probabilities $P(z < -z_0)$ \begin{equation} P( z < -\beta) = \frac{1-P( -\beta < z < \beta )}{2} = 0.35 \end{equation} Referring to the z-table, we get that \begin{equation} \beta = 0.1 \sqrt{n} = 1.81 \end{equation} so \begin{equation} \sqrt{n} = \frac{1.81}{0.1} = 18.1 \end{equation} So $n = 328$.