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Mkey [24]
1 year ago
8

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?
Mathematics
1 answer:
Elis [28]1 year ago
7 0

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

<h3>Normal Probability Distribution</h3>

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 <u>one subtracted by the p-value of Z when X = 89.5</u>, 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 brainly.com/question/15181104

#SPJ1

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lisabon 2012 [21]

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Step-by-step explanation:

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See Below

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Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t
andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ.e^{-\lambda.x}

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - e^{-\lambda.x}

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F(100) = 1 - e^{-0.0143.100}

F(100) = 0.76

<u>Probability</u> of distance at most <u>200</u>:

F(200) = 1 - e^{-0.0143.200}

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F(100≤X≤200) = 0.18

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E(X) = \frac{1}{\lambda}

E(X) = \frac{1}{0.0143}

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

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(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

\int\limits^m_0 f({x}) \, dx = 0.5

\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx = 0.5

\lambda.\frac{e^{-\lambda.x}}{-\lambda} = -e^{-\lambda.x} + e^{0}

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-0.0143.m = - 0.0693

m = 48.46

6 0
3 years ago
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