Question

A car is known to be 88% likely to pass inspection at a certain motor vehicle agency inspection office. what is the probability that at least 90 cars pass inspection if a random sample of 100 cars is taken at this motor vehicle agency inspection office?

Answer 1

Using the normal distribution, there is a 0.3228 = 32.28% probability that at least 90 cars pass inspection.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$.

The parameters for the binomial distribution are given by:

n = 100, p = 0.88.

Hence the mean and the standard deviation for the approximation are:

• $\mu = np = 100 \times 0.88 = 88$

• $\sigma = \sqrt{np(1-p)} = \sqrt{100 \times 0.88 \times 0.12} = 3.25$

The probability that at least 90 cars pass inspection, using continuity correction, is P(X > 89.5), which is one subtracted by the p-value of Z when X = 89.5, hence:

Z = (89.5 - 88)/3.25

Z = 0.46

Z = 0.46 has a p-value of 0.6772.

1 - 0.6772 = 0.3228.

0.3228 = 32.28% probability that at least 90 cars pass inspection.

More can be learned about the normal distribution at https://brainly.com/question/15181104

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