Question

62(b) A continuous manufacturing process produces items whose weights are normally distributed withmean weight of 800 grams and standard deviation of 300 grams. A random sample of 36 items isto be drawn from the process. What is the probability that the arithmetic mean of the sample liesbetween 900 grams and 950 grams? Interpret the result.​

Answer 1

SOLUTION

GIVEN

A continuous manufacturing process produces items whose weights are normally distributed with mean weight of 800 grams and standard deviation of 300 grams.

A random sample of 36 items is to be drawn from the process.

TO DETERMINE

The probability that the arithmetic mean of the sample lies between 900 grams and 950 grams

EVALUATION

Here it is given that mean weight is 800 grams and standard deviation is 300 grams.

Also a random sample of 36 items is to be drawn from the process.

Thus we get

$\sf{n = 36 \:, \: \mu = 800 \: , \: \sigma = 300}$

$\sf{ \therefore \: \:S.E \: \: of \: \: \bar{x}}$

$\displaystyle\sf{ = \frac{ \sigma}{ \sqrt{n} } }$

$\displaystyle\sf{ = \frac{300}{ \sqrt{36} } }$

$\displaystyle\sf{ = \frac{300}{ 6 } }$

$\displaystyle\sf{ =50 }$

$\displaystyle\sf{ \therefore \: \:z = \frac{ \bar{x} - \mu}{S.E \: \: of \: \: \bar{x}} }$

$\displaystyle\sf{ \therefore \: \:z = \frac{ \bar{x} - 800}{50} }$

$\displaystyle\sf{ When \: \bar{x} = 900 \: \: we \: have \: \:z = \frac{ 900 - 800}{50} = 2}$

$\displaystyle\sf{ When \: \bar{x} = 950 \: \: we \: have \: \:z = \frac{ 950 - 800}{50} = 3}$

Hence the required probability that the arithmetic mean of the sample lies between 900 grams and 950 grams

$\sf{P(900 < \bar{x}< 950)}$

$\sf{ = P(2 < z< 3)}$

$\sf{ = 0.4987 - 0.4772}$

$\sf{ = 0.0215 \: }$

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