Question

Assume that 90% of all industries contain shipping open order files in their computerized database. In a random sample of 10 industries, let X be the number that includes shipping open order files in their computerized database.
i) Find P(X=8)
ii) Find P(X>5)
iii) Find the mean and variance of X.

Answer 1

Answer on Question #42119 - Math - Statistics and Probability

Assume that $90\%$ of all industries contain shipping open order files in their computerized database. In a random sample of 10 industries, let $X$ be the number that includes shipping open order files in their computerized database.

i) Find $P(X = 8)$

ii) Find $P(X > 5)$

iii) Find the mean and variance of $X$.

Solution.

We have the Bernoulli scheme with $n = 10$ trials.

Let $p$ is success probability. We've got that $p = 0.9$.

Now, we construct the probability function:

$$f(k) = P(X = k) = C_{10}^{k} \cdot p^{k} \cdot (1 - p)^{10 - k}$$

i) $P(X = 8) = C_{10}^{8} \cdot p^{8} \cdot (1 - p)^{8} = \frac{10!}{8! \cdot 2!} \cdot 0.9^{8} \cdot 0.1^{2} \approx 0.1937$

ii) $P(X > 5) = \sum_{k=6}^{10} C_{10}^{k} \cdot p^{k} \cdot (1 - p)^{10 - k} \approx 0.9984$

iii) It is Bernoulli scheme, so we have binomial distribution. Then, mean of $X$ is equal to $n \cdot p = 10 \cdot 0.9 = 9$. Variance of $X$ is equal to $n \cdot p \cdot (1 - p) = 10 \cdot 0.9 \cdot 0.1 = 0.9$

Answer: i) $P(X = 8) \approx 0.1937$,

ii) $P(X > 5) \approx 0.9984$,

iv) Mean of $X$ is 9, variance of $X$ is 0.9.

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