Question

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 10%. Let X = the number of defective boards in a random sample of size n = 20, so X ~ Bin(20, 0.10). (Round your probabilities to three decimal places.)(a) Determine P(X ≤ 2).(b) Determine P(X ≥ 5).(c) Determine P(1 ≤ X ≤ 4).(d) What is the probability that none of the 25 boards is defective?(e) Calculate the expected value and standard deviation of X. (Round your standard deviation to two decimal places.)expected value boardsstandard deviation boards

Answer 1

a) P(X ≤ 2) = 0.677

b) P(X ≥ 5) = 0.043

c) P(1 ≤ X ≤ 4) = 0.835

d) P(x=0) = 0.072

e) expected value: 2 boards

standard deviation: 1.34 boards

Explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.1.

The probability that k number of defective boards in the sample is:
`P(x=k)=\dbinom{n}{k}p^k(1-p)^(n-k)=\dbinom{20}{k}\cdot0.1^k\cdot0.9^(20-k)`

a) We have to calculate the probability that 2 or less number of defective boards. This can be calculated as:

`P(x\leq2)=P(x=0)+P(x=1)+P(x=2)\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.1^(0)\cdot0.9^(20)=1\cdot1\cdot0.122=0.122\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.1^(1)\cdot0.9^(19)=20\cdot0.1\cdot0.135=0.270\\\\\\P(x=2)=\dbinom{20}{2}\cdot0.1^(2)\cdot0.9^(18)=190\cdot0.01\cdot0.15=0.285\\\\\\\\P(x\leq2)=0.122+0.270+0.285\\\\\\P(x\leq2)=0.677`

The probability that 2 or less number of defective boards is 0.677.

b) We have to calculate the probability that 5 or more number of defective boards. This can be calculated as:

`P(x\geq5)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.1^(0)\cdot0.9^(20)=1\cdot1\cdot0.122=0.122\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.1^(1)\cdot0.9^(19)=20\cdot0.1\cdot0.135=0.270\\\\\\P(x=2)=\dbinom{20}{2}\cdot0.1^(2)\cdot0.9^(18)=190\cdot0.01\cdot0.15=0.285\\\\\\P(x=3)=\dbinom{20}{3}\cdot0.1^(3)\cdot0.9^(17)=1140\cdot0.001\cdot0.167=0.190\\\\\\P(x=4)=\dbinom{20}{4}\cdot0.1^(4)\cdot0.9^(16)=4845\cdot0.0001\cdot0.185=0.090\\\\\\\\\\\\`

`P(x\geq5)=1-[0.122+0.270+0.285+0.190+0.090]\\\\P(x\geq5)=1-0.957=0.043`

The probability that 5 or more number of defective boards is 0.043.

c) We have to calculate P(1 ≤ X ≤ 4)

`P(1\leq X \leq4)=P(x=1)+P(x=2)+P(x=3)+P(x=4)\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.1^(1)\cdot0.9^(19)=20\cdot0.1\cdot0.135=0.270\\\\\\P(x=2)=\dbinom{20}{2}\cdot0.1^(2)\cdot0.9^(18)=190\cdot0.01\cdot0.15=0.285\\\\\\P(x=3)=\dbinom{20}{3}\cdot0.1^(3)\cdot0.9^(17)=1140\cdot0.001\cdot0.167=0.190\\\\\\P(x=4)=\dbinom{20}{4}\cdot0.1^(4)\cdot0.9^(16)=4845\cdot0.0001\cdot0.185=0.090\\\\\\\\P(1\leq X \leq4)=0.270+0.285+0.190+0.090=0.835`

d) The probability that none of the 25 boards is defective is P(x=0), but now calculated for a sample size n=25.

`P(x=0)=\dbinom{25}{0}\cdot0.1^(0)\cdot0.9^(25)=1\cdot1\cdot0.072=0.072\\\\\\`

e) The expected value and standard deviation for X (n=20, p=0.1) are:

`\mu=np=20\cdot0.1=2\\\\\sigma=√(np(1-p))=√(20\cdot0.1\cdot0.9)=√(1.8)=1.34`