Chapter 6: Q.85 (page 393)

A NUMMI assembly line, which has been operating since $1984$ , has built an average of $6, 000$ cars and trucks a week. Generally, $10 %$ of the cars were defective coming off the assembly line. Suppose we draw a random sample of $n = 100$ cars. Let $X$ represent the number of defective cars in the sample. What can we say about $X$ in regard to the $68 - 95 - 99.7$ empirical rule (one standard deviation, two standard deviations and three standard deviations from the mean are being referred to)? Assume a normal distribution for the defective cars in the sample.

Short Answer

Expert verified

$68 %$ of the values of $X$ lie between $7$ and $13, 95 %$ of the values of $X$ lie between $4$ and $16$ , and $99.7 %$ of the values of $X$ lie between $2$ and $19$ .

Step by step solution

Given Information

The number of defective autos in a sample is represented by a regularly distributed random variable called $X$ .

The sample's car count, $n = 100$ .

Car defect probability, $% = 0.1, p = 10 %$ .

Explanation

Mean, $μ = n p$

$= 100 * 0.1$

$= 10$

Standard deviation, $σ = \sqrt{100 * 0.1 * (1 - 0.1)}$

$= 3$

i. For $z = \pm 1,$

$x_{1} = μ - z σ$

and

$x_{2} = μ + z σ$

$\Rightarrow x_{1} = 10 - 1 * 3$

$= 7$

$x_{2} = 10 + 1 * 3$

$= 13$

$X$ values range from $7$ to $13$ in $68 %$ of cases.

Explanation

ii. For $z = \pm 2,$

$x_{1} = μ - z σ$

and

x_{2} = μ + z σ

$\Rightarrow x_{1} = 10 - 2 * 3$

$= 4$

and

$x_{2} = 10 + 2 * 3$

$= 16$

$95 %$ of $X$ values are between $4$ and $16$ .

iii. For $z = \pm 3,$

$x_{1} = μ - z σ$

and

x_{2} = μ + z σ

\Rightarrow x_{1} = 10 - 3 * 3

$= 1$

x_{2} = 10 + 3 * 3

$= 19$

$99.7 %$ of the values of $X$ lie between $1$ and $19$ .

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