Question

) To estimate weight of a certain product, 100 of these products are measured. The average weight and standard deviation are calculated to be 40 lb and 1 lb, respectively. 10,000 products are produced every day and assume normal distribution. (a) (8) We have to reject any product that weighs less than 38 lb. About how many products are rejected every day? (b) (10) How many products are expected to weight between 39 lb and 42 lb? (c) (7) What is the 90% confidence interval on the mean weight?

Answer 1

a)
`P(X<38)=P((X-\mu)/(\sigma)<(38-\mu)/(\sigma))=P(Z<(38-40)/(1))=P(z<-2)`

And we can find this probability with the normal standard table or excel and we got:

`P(z<-2)=0.02275`

And we would expect about 0.02275*10000 =227.5 rejected and 2.75 in the sample of 100 selected

b)
`P(39<X<42)=P((39-\mu)/(\sigma)<(X-\mu)/(\sigma)<(42-\mu)/(\sigma))=P((39-40)/(1)<Z<(42-40)/(1))=P(-1<z<2)`

And we can find this probability with this difference:

`P(-1<z<2)=P(z<2)-P(z<-1)`

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

`P(-1<z<2)=P(z<2)-P(z<-1)=0.97725-0.159=0.819`

And we expect about 0.819*10000= 8190 and 81.9 in the sample od 100 selected

c)
`40-1.64(1)/(√(100))=39.836`

`40+1.64(1)/(√(100))=40.164`

So on this case the 90% confidence interval would be given by (39.836;40.164)

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".

Part a

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(40,1)`

Where
`\mu=40` and
`\sigma=1`

We are interested on this probability

`P(X<38)`

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

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

If we apply this formula to our probability we got this:

`P(X<38)=P((X-\mu)/(\sigma)<(38-\mu)/(\sigma))=P(Z<(38-40)/(1))=P(z<-2)`

And we can find this probability with the normal standard table or excel and we got:

`P(z<-2)=0.02275`

And we would expect about 0.02275*10000 =227.5 rejected and 2.75 in the sample of 100 selected

Part b

`P(39<X<42)=P((39-\mu)/(\sigma)<(X-\mu)/(\sigma)<(42-\mu)/(\sigma))=P((39-40)/(1)<Z<(42-40)/(1))=P(-1<z<2)`

And we can find this probability with this difference:

`P(-1<z<2)=P(z<2)-P(z<-1)`

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

`P(-1<z<2)=P(z<2)-P(z<-1)=0.97725-0.159=0.819`

And we expect about 0.819*10000= 8190 and 81.9 in the sample od 100 selected

Part c

The confidence interval for the mean is given by the following formula:

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that
`z_(\alpha/2)=1.64`

Now we have everything in order to replace into formula (1):

`40-1.64(1)/(√(100))=39.836`

`40+1.64(1)/(√(100))=40.164`

So on this case the 90% confidence interval would be given by (39.836;40.164)