Question

the mean weight for a part made using a new production process is 2.09 pounds. assume that a normal distribution applies and that the standard deviation is 0.21 pounds. Based on this paragraph of text, use the correct excel output above to answer the following question. Find the probability that weight for a part made using the new production process is less than 1.61 or greater than 2.53 pounds

Answer 1

`P(X<1.61) = P(z<(1.61-2.09)/(0.21)) = P(Z<-2.286)=0.011`

`P(X>2.53) = P(z>(2.53-2.09)/(0.21)) = P(Z>2.095)=1-P(Z<2.095)= 1-0.982= 0.018`

And adding the probabilites we got:

`P(X< 1.61 \cup X>2.53)= 0.011+ 0.018 = 0.029`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

`X \sim N(2.09,0.21)`

Where
`\mu=2.09` and
`\sigma=0.21`

We are interested on this probability

`P(X< 1.61 \cup X>2.53)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

We can find the individual probabilites like this:

`P(X<1.61) = P(z<(1.61-2.09)/(0.21)) = P(Z<-2.286)=0.011`

`P(X>2.53) = P(z>(2.53-2.09)/(0.21)) = P(Z>2.095)=1-P(Z<2.095)= 1-0.982= 0.018`

And adding the probabilites we got:

`P(X< 1.61 \cup X>2.53)= 0.011+ 0.018 = 0.029`

Answer 2

2.89% probability that weight for a part made using the new production process is less than 1.61 or greater than 2.53 pounds

Explanation:

Problems of normally distributed samples are 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 = (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.

In this problem, we have that:

`\mu = 2.09, \sigma = 0.21`

Find the probability that weight for a part made using the new production process is less than 1.61 or greater than 2.53 pounds

Less than 1.61

This is the pvalue of Z when X = 1.61. So

`Z = (X - \mu)/(\sigma)`

`Z = (1.61 - 2.09)/(0.21)`

`Z = -2.29`

`Z = -2.29` has a pvalue of 0.0110

Greater than 2.53

1 subtracted by the pvalue of Z when X = 2.53.

`Z = (X - \mu)/(\sigma)`

`Z = (2.53 - 2.09)/(0.21)`

`Z = 2.1`

`Z = 2.1` has a pvalue of 0.9821

1 - 0.9821 = 0.0179

Less than 1.61 or greater than 2.53 pounds

0.0110 + 0.0179 = 0.0289

2.89% probability that weight for a part made using the new production process is less than 1.61 or greater than 2.53 pounds