Random variable X has a distribution: P(X=0)=0.2;...

Random variable X has a distribution: P(X=0)=0.2; P(X=1)=0.3; P(X=2)=0.4; P(X=3) P(X=4)=0.1. Find a) E(X) & Var(X). 0.3 b) Find $F_X(x_0)$ $\leftarrow$ cumulative distribution c) Find quantile of order $\frac{1}{4}$ & median d) Find P(2 $\le$ X $\le$ 4) (ii) For sunflower seed the power of sprouting is 92%. We choose 200 seeds. Find the probability that $X_{200} = X_1 + ... + X_{200}$ germinated seeds is $186 \le X_{200} \le 190$

Transcript

00:01 So in this case we want to calculate variance from the given probability distribution.

00:09 So the first step is to find the mean.

00:12 And to get the mean, we say x values times their probability distribution.

00:19 So 1 times 0 .1, it's 0 .1.

00:23 And then 2 times 0 .5, it's 1 .0.

00:28 And then 3 times 0 .2 it's 0 .6 and 4 times 0 .2 it's 0 .8.

00:42 Now we have to sum that.

00:46 So to sum that, let me put the sum column here where sums are required.

00:53 So to sum this 0 .1 plus 1 .0 .0 .6 plus 0 .8.

00:59 We get 2 .5.

01:04 Now 2 .5 is our mean for the distribution.

01:09 Now, in line with the formula for variance, we are supposed now to subtract, we are subtracting the mean from x.

01:24 So it will be for the first line on the, on the row, the next column.

01:33 We say 1 minus 2 .5 it's a 0 point or negative 0, in fact not negative 0 but it's it's negative 1 .5.

01:55 So we also see 2 minus 2 .5 it's negative 0 .5 then 3 minus 2 .5 it's 0 .5.

02:11 And then 4 minus 2 .5, it's 1 .5.

02:19 Okay, so that step is done.

02:22 If you add that column, you'll get a zero, so there's no need for us doing just that.

02:29 So the next step now is to square those the variances.

02:39 So we can square them by saying negative.

02:43 1 .5 squared it's 2 .25 then negative 0 .5 squared 0 .25 .0 .5 squared 0 .25 squared.

03:05 1 .5 squared...

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