If the probability mass function (p.m.f.) of a random variable $X$ is given by the table below,then find the value of $k$.
$X$ $1$ $2$ $3$ $4$ $5$ $P(X = x)$ $k$ $\frac{k}{3}$ $\frac{k}{4}$ $\frac{k}{2}$ $\frac{k}{2}$
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $P(X = x)$ | $k$ | $\frac{k}{3}$ | $\frac{k}{4}$ | $\frac{k}{2}$ | $\frac{k}{2}$ |
- A$\frac{15}{31}$
- B$\frac{1}{12}$
- C$\frac{11}{12}$
- D$\frac{12}{31}$
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If a random variable $X$ has the following probability distribution, then its variance is $X=x$ $1$ $3$ $5$ $2$ $P(X=x)$ $3 K^2$ $K$ $K^2$ $2 K$
| $X=x$ | $1$ | $3$ | $5$ | $2$ |
| $P(X=x)$ | $3 K^2$ | $K$ | $K^2$ | $2 K$ |
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If $X$ is a random variable with the following probability distribution:
$X=x$ $-3$ $6$ $9$ $P(X=x)$ $\frac{1}{6}$ $\frac{1}{2}$ $\frac{1}{3}$
Then the variance of $X$ is:
| $X=x$ | $-3$ | $6$ | $9$ |
| $P(X=x)$ | $\frac{1}{6}$ | $\frac{1}{2}$ | $\frac{1}{3}$ |
Then the variance of $X$ is:
State whether the following table represents a probability distribution of a random variable $X$. Give reasons for your answer.
$X$ $0$ $1$ $2$ $P(X)$ $0.4$ $0.4$ $0.2$
| $X$ | $0$ | $1$ | $2$ |
| $P(X)$ | $0.4$ | $0.4$ | $0.2$ |
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View SolutionIn a hospital, on an average if there are $35$ births in a week, then the probability that there will be less than $3$ births in a day is:
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