Mathematics, 26.02.2020 18:59 vrentadrienneoqug1a

A random sample of 1,700 cars is checked for a serious defect. The chance that a car has the defect is 0.6%. Approximate the probability that at least 10 cars in the sample have the defect. State and check the appropriate conditions.

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Answer from: tburlew01

52.39% probability that at least 10 cars in the sample have the defect.

Conditions:

Each car has only two possible outcomes. Either it is defective, or it is not. The cars are chosen at random, which means that the probability of a car being defective is independent from other cars. So we use the binomial probability distribution to solve this questions.

Step-by-step explanation:

Conditions:

We are working with a large sample, which means that we can approximate the binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

n = 1700, p = 0.006

E(X) = np = 1700*0.006 = 10.2

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1700*0.006*0.994} = 3.18$

Approximate the probability that at least 10 cars in the sample have the defect.

This is 1 subtracted by the pvalue of Z when X = 10. So

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

$Z = \frac{9 - 10.2}{3.18}$

Z = -0.06

Z = -0.06 has a pvalue of 0.4761.

1 - 0.4761 = 0.5239

52.39% probability that at least 10 cars in the sample have the defect.

Answer from: Quest

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like, however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

$\bf e(\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-2}{)}\implies \implies \cfrac{-2}{0+2}\implies \cfrac{-2}{2}\implies -1 \\\\[-0.35em] ~\dotfill$

$\bf d(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{)}{)}\implies \cfrac{0+1}{0+2}\implies \cfrac{1}{2} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope~of~ef}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}}$

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

Answer from: Quest

351°f

step-by-step explanation:

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A random sample of 1,700 cars is checked for a serious defect. The chance that a car has the defect...