The number of defective items in a production line follows a binomial distribution with parameters n =10 and p=0.2 a) find probability that exactly 3 items are defective b) calculate the mean and variance of the number of defective items

Economics

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Answers

2 parts
a

0.201326592≈0.20130.201326592 \approx 0.2013

b
2
1.6

Step-by-step solution

4 steps

Here's the solution for your question:

The number of defective items in a production line follows a binomial distribution with parameters n=10n=10 and p=0.2p=0.2.
Let XX be the number of defective items. So, X∼B(10,0.2)X \sim B(10, 0.2).
The probability mass function for a binomial distribution is P(X=k)=(nk)pk(1−p)n−kP(X=k) = \binom{n}{k} p^k (1-p)^{n-k}.

Part a· 2 steps

Find the probability that exactly 3 items are defective.

  1. Identify the parameters

    n=10n=10, p=0.2p=0.2, and k=3k=3.
    The probability of success (defective item) is p=0.2p=0.2.
    The probability of failure (non-defective item) is 1−p=1−0.2=0.81-p = 1-0.2 = 0.8.

  2. Apply the binomial probability formula.

    P(X=3)=(103)(0.2)3(0.8)10−3P(X=3) = \binom{10}{3} (0.2)^3 (0.8)^{10-3} P(X=3)=10!3!(10−3)!(0.2)3(0.8)7P(X=3) = \frac{10!}{3!(10-3)!} (0.2)^3 (0.8)^7 P(X=3)=10×9×83×2×1×(0.008)×(0.2097152)P(X=3) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times (0.008) \times (0.2097152) P(X=3)=120×0.008×0.2097152P(X=3) = 120 \times 0.008 \times 0.2097152 P(X=3)=0.96×0.2097152P(X=3) = 0.96 \times 0.2097152 P(X=3)=0.201326592≈0.2013P(X=3) = \boxed{0.201326592 \approx 0.2013}

Part b2

Calculate the mean and variance of the number of defective items.

  1. Calculate the mean E(X)E(X).

    For a binomial distribution, the mean is given by E(X)=npE(X) = np.

    E(X)=10×0.2E(X) = 10 \times 0.2 E(X)=2E(X) = \boxed{2}

  2. Calculate the variance Var(X)\text{Var}(X).

    For a binomial distribution, the variance is given by Var(X)=np(1−p)\text{Var}(X) = np(1-p).

    Var(X)=10×0.2×(1−0.2)\text{Var}(X) = 10 \times 0.2 \times (1 - 0.2) Var(X)=10×0.2×0.8\text{Var}(X) = 10 \times 0.2 \times 0.8 Var(X)=2×0.8\text{Var}(X) = 2 \times 0.8 Var(X)=1.6\text{Var}(X) = \boxed{1.6}
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