Question

Add a cell phone assembly plant, 80% of the cell phone keypad pass inspection. A random sample of 181 keypads is analyzed. Find the probability that less than 76 in the sample keypads pass inspection.

Answer 1

0% probability that less than 76 in the sample keypads pass inspection.

Explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

In this problem, we have that:

`n = 181, p = 0.8`

So

`\mu = E(X) = np = 181*0.8 = 144.8`

`√(V(X)) = √(np(1-p)) = √(181*0.8*0.2) = 5.38`

Find the probability that less than 76 in the sample keypads pass inspection.

Using continuity correction, this is
`P(X < 76 - 0.5) = P(X < 75.5)`, which is the pvalue of Z when X = 75.5.

So

`Z = (X - \mu)/(\sigma)`

`Z = (75.5 - 144.8)/(5.38)`

`Z = -12.88`

`Z = -12.88` has a pvalue of 0

0% probability that less than 76 in the sample keypads pass inspection.