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A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.
Given: The class of 15 students with their ages.
Form the given information we can draw a table:
X | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |
f | 2 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |
P(X = 14)
P(X = 15)
P(X = 16)
P(X = 17)
P(X = 18)
P(X = 19)
P(X = 20)
P(X = 21)
Hence, the required probability distribution is,
X | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |
P(X) |
Therefore E(X) is:
⇒ E(X) = 17.53
And E(X2) is:
⇒ E(X2) = 312.2
Then Variance, Var(X) = E(X2) – (E(X))2
= 312.2 – (17.53)2
= 312.2 – 307.417 ≈ 4.78
And Standard deviation
⇒ Standard deviation ≈ 2.19
A random variable X has the following probability distribution:
X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
P(X) | 0 | k | 2k | 2k | 3k | K2 | 2 K2 | 7 K2 + k |
Determine
(i) k (ii) P(X < 3)
(iii) P(X > 6) (iv) P(0 < X < 3)