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Sagot :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].
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:
- [tex]\mu = np = 100 \times 0.88 = 88[/tex]
- [tex]\sigma = \sqrt{np(1-p)} = \sqrt{100 \times 0.88 \times 0.12} = 3.25[/tex]
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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