How can I change uniform to normal?

Question

I'm taking one statistic course, and I'm confused...

the question is

A candymaker produces mints that have a label weight of 10 grams. Assume that the distribution of the weight of such a mint is uniform over (9.5, 10.5).

(a) Let X be the weight of a single mint selected at random from the production line. Find P(9.9 < X < 10.1)

So what I can't understand is, "unifrom over (9.5, 10.5)"

Here, is my solution, but I'm not sure it is right or wrong..

I think, E of uniform is (a+b)/2 and V of unifrom is (b-a)^2/12

so, I assume that if I change it to normal distribution, I would get N(10, 1/12).

Then, P( (9.9 - 10)/root(1/12) < Z < (10.1 - 10)/root(1/12))

=> P(-0.35 < Z <0.35). So, P(Z<0.35) = 0.6368, then P(-0.35 < Z < 0.35) is 0.2736.

However, professor gave answer sheet, and it said it is "0.2"(there's no solution, just answer)

What part am I wrong at? or it is just professor think 0.6368 ~ 0.6?

If professor says P(Z<0.35) ~ 0.6, then 0.2 might be correct.

But I'm not sure what is right.

If my solution is wrong, can you tell me what is wrong and explain it?(like, uniform to normal is wrong)

Answer 1

Don't convert to a normal distribution unless you are looking at the sum or the average of many uniformly distributed random variables.

Uniform $p(x) = \begin{cases}\frac {1}{b-a}&x\in (a,b)\\0&x\notin(a,b)\end{cases}$

When $x$ is uniformly distributed as described above.

$P(x\in (9.9, 10.1)) = \frac {10.1-9.9}{10.5-9.5} = 0.2$

Answer 2

Uniform distribution is something entirely different than normal distribution, so you should not try to switch to normal distribution (whatever that means).

Uniform distribution means that the probability is constant over the interval. In particular, if something is uniform over an interval $[a,b]$ it means that the probability density function is $$f(x) = \begin{cases} 1/(b-a) &\mbox{if } a\leq x\leq b \\ 0 & \mbox{otherwise } \end{cases} $$ Then you have that $$P(x_1 < X < x_2) = \int_{x_1}^{x_2} f(x)\, \mathrm d x.$$