$x$ 0 1 2 3 4 5 6 7 P(x) 0 2k k 3k $2k^2$ 2k $k^2+k$ $7k^2$
then $P(3 < x \le 6)$ is equal to
Determine the value of constant k
The sum of probabilities for all possible values of a random variable must equal 1.
Given the probability distribution:
- $P(x=0) = 0$
- $P(x=1) = 2k$
- $P(x=2) = k$
- $P(x=3) = 3k$
- $P(x=4) = 2k^2$
- $P(x=5) = 2k$
- $P(x=6) = k^2+k$
- $P(x=7) = 7k^2$
Summing these probabilities gives:
$ \sum_{x=0}^{7} P(x) = 0 + 2k + k + 3k + 2k^2 + 2k + k^2 + k + 7k^2 $ $ \sum P(x) = (2+1+3+2+1)k + (2+1+7)k^2 $ $ \sum P(x) = 9k + 10k^2 $Set the sum equal to 1:
$ 10k^2 + 9k = 1 $Rearrange into a standard quadratic equation:
$ 10k^2 + 9k - 1 = 0 $Factor the quadratic equation:
$ (10k - 1)(k + 1) = 0 $This gives two possible values for $k$: $k = \frac{1}{10}$ or $k = -1$. Since probabilities must be non-negative, $k$ must be positive.
Therefore, $k = \frac{1}{10} = 0.1$.
Calculate the probability $P(3 < x \le 6)$
The required probability $P(3 < x \le 6)$ includes the values $x=4$, $x=5$, and $x=6$.
Calculate the probabilities for these values:
- $P(x=4) = 2k^2 = 2(0.1)^2 = 2(0.01) = 0.02$
- $P(x=5) = 2k = 2(0.1) = 0.2$
- $P(x=6) = k^2+k = (0.1)^2 + 0.1 = 0.01 + 0.1 = 0.11$
Sum these probabilities:
$ P(3 < x \le 6) = P(x=4) + P(x=5) + P(x=6) $ $ P(3 < x \le 6) = 0.02 + 0.2 + 0.11 $ $ P(3 < x \le 6) = 0.33 $Similar Questions
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