Question

At a cell phone assembly plant, 75% of the cell phone keypads pass inspection. A random sample of 110 keypads is analyzed. Find the probability that more than 78% of the sample keypads pass inspection. Use at least five decimal places for the denominator.

Answer 1

23.27%

Explanation:

From the statement we know that random sample n is 110 and that p is 75% and x the percentage to evaluate is 78%

We have that the probability would be equal:

P (x > 0.78) =
`P(z <\frac{x-p}{\sqrt{(p*(1-p))/(n) }})`

Replacing we have:

`P(z <\frac{0.78-0.75}{\sqrt{(0.75*(1-0.75))/(110) }})`

P ( z < 0.73) = 1 - P ( z => 0.73)

= 1 - 0.7673

= 0.2327

Therefore the probability is 23.27%