Calculating the minimum size of a sample

A company has 2.400 employees. You want to extract a sample of employees to test the quality of the food of the refectory. On a pilot sample, in a scale of 1 to 10, the satisfaction degree grades got a mean of $6.5$ and a SD of $.2.0$

a) Calculate the minimum size of the sample, with an maximum error of $0.5$ and a confidence level of $99\%$.

b) Given the size of the sample, suppose that $70$ employees gave a grade equal or greater than $5.0$. Show a confidence interval with $90\%$ confidence level for the percentage of the individuals that attributed grades equal or greater than $5.0$.

I can't get past a). My book says the answer is $102$, but calculating $$n \geq \frac{z_0^2\sigma^2}{E_0^2} = \frac{2,576^2*2^2}{0,5^2} \approx106.1724 $$

I don't get why this is wrong, the SD and max error can't be other. I'm suspecting of $z_o$, but t-distribution is giving me $2,576$ for $99\%$. Also, I don't have any ideas of how should I start, neither the prccesses to solve b).

Any tips?

Solution 1:

Your answer is correct.

Thus the minimum size is $n=107$. Perhaps there is an error in the book's solution

For b) set the sample mean $\overline{X}=\frac{70}{107}$ and use the confidence intervals for the proportion.

If you want to use the book a) result, set $\overline{X}=\frac{70}{102}$,