Question

2.A production process manufactures items with weights that are normally distributed with mean 10 pounds and standard deviation 0.1. An item is considered to be defective if its weight is less than 9.8 pounds or greater than 10.2 pounds. Suppose that these items are currently produced in batches of 1000 units. a.Find the probability that at most 5% of the items in a given batch will be defective b.Find the probability that at least 85% of these items in a given batch will be acceptable.

Answer 1

Answer:

Step-by-step explanation:

Given that:

population mean = 10

standard deviation = 0.1

sample mean = 9.8 < x > 10.2

The z score can be computed as:

$z = \dfrac{\bar x - \mu}{\sigma}$

if x > 10.2

$z = \dfrac{10.2- 10}{0.1}$

$z = \dfrac{0.2}{0.1}$

z = 2

If x < 9.8

$z = \dfrac{9.8- 10}{0.1}$

$z = \dfrac{-0.2}{0.1}$

z = -2

The p-value = P (z ≤ 2) + P (z ≥ 2)

The p-value = P (z ≤ 2) + ( 1 -  P (z ≥ 2)

p-value = 0.022750 +(1 -   0.97725)

p-value = 0.022750 +  0.022750

p-value = 0.0455

Therefore; the probability of defectives  = 4.55%

the probability of acceptable = 1 - the probability of defectives

the probability of acceptable = 1 - 0.0455

the probability of acceptable = 0.9545

the probability of acceptable = 95.45%

4.55% are defective or 95.45% is acceptable.

sampling distribution of proportions:

sample size n=1000

p = 0.0455

The z - score for this distribution at most 5% of the items is;

$z = \dfrac{0.05 - 0.0455}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}$

$z = \dfrac{0.0045}{\sqrt{\dfrac{0.04342975}{1000}}}$

$z = \dfrac{0.0045}{\sqrt{4.342975 \times 10^{-5}}}$

z = 0.6828

The p-value = P(z ≤ 0.6828)

From the z tables

p-value = 0.7526

Thus, the probability that at most 5% of the items in a given batch will be defective = 0.7526

The z - score for this distribution for at least 85% of the items is;

$z = \dfrac{0.85 - 0.9545}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}$

$z = \dfrac{-0.1045}{\sqrt{\dfrac{0.04342975}{1000}}}$

z = −15.86

p-value = P(z ≥  -15.86)

p-value = 1 - P(z <  -15.86)

p-value = 1 - 0

p-value = 1

Thus, the probability that at least 85% of these items in a given batch will be acceptable = 1

Answer 2

a. 0.7526

b. 1

Given that:

$\sigma = 0.1\\\mu = 10$

It is given that item is acceptable if 9.8 < X < 10.2

Calculating standard variate value for :

1. x = 9.8:

$z = \dfrac{x - \mu}{\sigma}$

z = -2

2. x=  10.2

$z = \dfrac{x - \mu}{\sigma}$

z = 2

Now we can convert P( 9.8 < X < 10.2) to P(-2 < Z < 2).

Calculating

P(-2 < Z < 2) = P(Z <2) - P(Z < - 2)

                   = 0.9772 - 0.0227           (refer these values from z score table)

                   =  0.9545

Thus, the probability that an item from that population if randomly selected, it has 0.9545 probability to be acceptable.

And thus probability of an item to be defective is 1 - 0.9545 = 0.0455.

The batch size is 1000 unit.

The probability that at most 5% items are defective is calculated by z scores:

$z = \dfrac{0.05-0.045}{\sqrt[2]{\dfrac{0.0455\times 0.9545}{1000}}}$

or

z = 0.6828

P( at most 5% defective) =    $P(Z \leq 0.6828)$

= 0.7526 ( from z score table).

$z = \dfrac{0.85-0.9545}{\sqrt[2]{\dfrac{0.0455\times 0.9545}{1000}}}$

or

z = -15.8

and thus

P( at least 85% acceptable ) = P(Z >= -15.8) = 1- P(Z < -15.8)

P( at least 85% acceptable ) = 1 - 0 = 1

The z score tables give p values.

For more information, refer to:

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