Question
Two numbers are selected at random (without replacement) from positive integers 2, 3, 4, 5, 6 and 7. Let X denote the larger of the two numbers obtained. Find the mean and variance of the probability distribution of X.
P(X = 3) = P(larger number is 3) = {(2, 3), (3, 2)} $=\frac{2}{30}$
P(X = 4) = P(larger number is 4) = {(2, 4), (4, 2), (3, 4), (4, 3)} $=\frac{4}{30}$
P(X = 5) = P(larger number is 5) = {(2, 5), (5, 2), (3, 5), (5, 3), (4, 5), (5, 4)} $=\frac{6}{30}$
P(X = 6) = P(larger number is 6) = {(2, 6), (6, 2), (3, 6), (6, 3), (4, 6), (6, 4), (5, 6), (6, 5)} $=\frac{8}{30}$
P(X = 7) = P(larger number is 7) = {(2, 7), (7, 2), (3, 7), (7, 3), (4, 7), (7, 4), (5, 7), (7, 5), (6, 7), (7, 6)} $=\frac{10}{30}$
Thus, the probability distribution of random variable X is,
| $\text{x}_\text{i}$ | $\text{p}_\text{i}$ | $\text{x}_\text{i}\text{p}_\text{i}$ | $\text{x}_\text{i}^2\text{p}_\text{i}$ |
| $3$ | $\frac{2}{30}$ | $\frac{6}{30}$ | $\frac{18}{30}$ |
| $4$ | $\frac{4}{30}$ | $\frac{16}{30}$ | $\frac{64}{30}$ |
| $5$ | $\frac{6}{30}$ | $\frac{30}{30}$ | $\frac{150}{30}$ |
| $6$ | $\frac{8}{30}$ | $\frac{48}{30}$ | $\frac{288}{30}$ |
| $7$ | $\frac{10}{30}$ | $\frac{70}{30}$ | $\frac{490}{30}$ |
|
|
| $\sum\text{x}_\text{i}\text{p}_\text{i}=\frac{17}{3}$ | $\sum\text{x}_\text{i}\text{p}_\text{i}^2=\frac{101}{3}$ |
Mean $=\sum\text{x}_\text{i}\text{p}_\text{i}=\frac{17}{3}$
Variance $=\sum\text{x}_\text{i}\text{p}_\text{i}-\big(\sum\text{x}_\text{i}\text{p}_\text{i}\big)^2=\frac{101}{3}-\Big(\frac{17}{3}\Big)=\frac{14}{9}$
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| Machine | Area occupied | Labour force | Daily output ( in units) |
| A | 1000m2 | 12 men | 60 |
| B | 1200m2 | 8 men | 40 |
He has
maximum area of 9000 $\text{m}^{2}$ available, and 72 skilled labourers who can operate both the machines. How many machines of each type should he buy to maximise the daily output?x - y + z = 2
2x - y = 0
2y - z = 1