Question

Discrete uniform distribution question. Photo of the question is attached. My working will be in the comments, please provide assistance.

Answer 1

I wanted to use this formula, but then i dot no account for the sample of 36.
\[P(2.1 \le X \le2.5) = \int\limits_{2.1}^{2.5} \frac{ 1 }{ 3 } dx\]

Answer 2

anything @samjordon

Answer 3

Lol this seems familiar to me ... just tryna jog my memory

Answer 4

Give me a couple of minutes

Answer 5

@perl you familiar with this?

Answer 6

you can use the distribution of sample means

Answer 7

It is true that
$$
\Large \overline X \sim N(\mu, \frac{\sigma}{\sqrt n})

$$

Answer 8

$$
\Large P( 2.1 \leq \overline X \leq 2.5 ) = P( \frac{2.1 - 2 }{\frac{\sqrt{2/3}}{6}} \leq Z \leq \frac{2.5 - 2 }{\frac{\sqrt{2/3}}{6}}) \\
=\Large .2310969
$$

I account for discrepancies in answers, due to rounding error.

Answer 9

so the first thing you have to do is find the mean.
thats pretty easy , it is 1*1/3 + 2*1/3 + 3*1/3.
now you have to find the standard deviation,

Answer 10

how did you get the standard deviation?

Answer 11

Mean and variance can be computed as follows:
\[\text{mean}=\mu=\sum_{i=1}^3x_if(x_i)\\
\text{variance}=\sigma^2=\sum_{i=1}^3{x_i}^2f(x_i)-\underbrace{\left(\sum_{i=1}^3x_if(x_i)\right)^2}_{\mu^2}\]
and standard deviation is the square root of the variance, i.e. \(\sigma\).

As @perl calculated above, the mean is
\[\mu=\sum_{i=1}^3x_if(x_i)=\frac{1}{3}+\frac{2}{3}+\frac{3}{3}\]
and the variance can be computed as
\[\sigma^2=\sum_{i=1}^3{x_i}^2f(x_i)-\left(\sum_{i=1}^3x_if(x_i)\right)^2=\frac{1}{3}+\frac{4}{3}+\frac{9}{3}-\mu^2~~\implies~~\sigma=\cdots\]

Answer 12

Thanks all.