Question

a locksmith claims that his lock can be forced open only 2% of the time, he samples 25 locks he placed during the last 6 months, what is the probability that more than 2.5% of locks were forced open?

Answer 1

The probability that more than 2.5% of locks were forced open = P(x > 0.025) = 0.42858

Explanation:

This is a normal distribution problem

Proportion = P = 2% = 0.02

Standard deviation of sample means = √[P(1-P)/n]

n = sample size = 25

σ = √[0.02(0.98)/25] = 0.028

The probability that more than 2.5% of locks were forced open = P(x > 0.025)

We first standardize 0.025

The standardized score of any value is the value minus the mean divided by the standard deviation.

z = (x - μ)/σ = (0.025 - 0.02)/0.028 = 0.18

To determine the probability that more than 2.5% of locks were forced open

P(x > 0.025) = P(z > 0.18)

We'll use data from the normal probability table for these probabilities

P(x > 0.025) = P(z > 0.18)

= 1 - P(z ≤ 0.18) = 1 - 0.57142 = 0.42858

Hope this Helps!!!