Question

4- A manufacturing process produces items whose weights are normally distributed. It is known that 22.57% of all the items produced weigh between 100 grams up to the mean and 49.18% weigh from the mean up to 190 grams. Determine the mean and the standard deviation.

Answer 1

Answer:

$\\ \mu = 118\;grams\;and\;\sigma=30\;grams$

Step-by-step explanation:

We need to use z-scores and a standard normal table to find the values that corresponds to the probabilities given, and then to solve a system of equations to find $\\ \mu\;and\;\sigma$.

First Case: items from 100 grams to the mean

For finding probabilities that corresponds to z-scores, we are going to use here a Standard Normal Table for cumulative probabilities from the mean (Standard normal table. Cumulative from the mean (0 to Z), 2020, in Wikipedia) that is, the "probability that a statistic is between 0 (the mean) and Z".

A value of a z-score for the probability P(100<x<mean) = 22.57% = 0.2257 corresponds to a value of z-score = 0.6, that is, the value is 0.6 standard deviations from the mean. Since this value is below the mean ("the items produced weigh between 100 grams up to the mean"), then the z-score is negative.

Then

$\\ z = -0.6\;and\;z = \frac{x-\mu}{\sigma}$

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

Second Case: items from the mean up to 190 grams

We can apply the same procedure as before. A value of a z-score for the probability P(mean<x<190) = 49.18% = 0.4918 corresponds to a value of z-score = 2.4, which is positive since it is after the mean.

Then

$\\ z =2.4\;and\; z = \frac{x-\mu}{\sigma}$

$\\ 2.4 = \frac{190-\mu}{\sigma}$ (2)

Solving a system of equations for values of the mean and standard deviation

Having equations (1) and (2), we can form a system of two equations and two unknowns values:

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

$\\ 2.4 = \frac{190-\mu}{\sigma}$ (2)

Rearranging these two equations:

$\\ -0.6*\sigma = 100-\mu$ (1)

$\\ 2.4*\sigma = 190-\mu$ (2)

To solve this system of equations, we can multiply (1) by -1, and them sum the two resulting equation:

$\\ 0.6*\sigma = -100+\mu$ (1)

$\\ 2.4*\sigma = 190-\mu$ (2)

Summing both equations, we obtain the following equation:

$\\ 3.0*\sigma = 90$

Then

$\\ \sigma = \frac{90}{3.0} = 30$

To find the value of the mean, we need to substitute the value obtained for the standard deviation in equation (2):

$\\ 2.4*30 = 190-\mu$ (2)

$\\ 2.4*30 - 190 = -\mu$

$\\ -2.4*30 + 190 = \mu$

$\\ \mu = 118$