Question

A particular fruit's weights are normally distributed, with a mean of 551 grams and a standard deviation of 20 grams. If you pick one fruit at random, what is the probability that it will weigh between 523 grams and 534 grams

Answer 1

`P(523<X<534)=P((523-\mu)/(\sigma)<(X-\mu)/(\sigma)<(534-\mu)/(\sigma))=P((523-551)/(20)<Z<(534-551)/(20))=P(-1.4<z<-0.85)`

And we can find this probability with this difference:

`P(-1.4<z<-0.85)=P(z<-0.85)-P(z<-1.4)`

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

`P(-1.4<z<-0.85)=P(z<-0.85)-P(z<-1.4)=0.198-0.0808=0.1172`

Explanation:

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

`X \sim N(551,20)`

Where
`\mu=551` and
`\sigma=20`

We are interested on this probability

`P(523<X<534)`

We can solve the 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(523<X<534)=P((523-\mu)/(\sigma)<(X-\mu)/(\sigma)<(534-\mu)/(\sigma))=P((523-551)/(20)<Z<(534-551)/(20))=P(-1.4<z<-0.85)`

And we can find this probability with this difference:

`P(-1.4<z<-0.85)=P(z<-0.85)-P(z<-1.4)`

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

`P(-1.4<z<-0.85)=P(z<-0.85)-P(z<-1.4)=0.198-0.0808=0.1172`

Answer 2

11.69% probability that it will weigh between 523 grams and 534 grams

Explanation:

Problems of normally distributed samples can be 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 = 551, \sigma = 20`

If you pick one fruit at random, what is the probability that it will weigh between 523 grams and 534 grams

This is the pvalue of Z when X = 534 subtracted by the pvalue of Z when X = 523. So

X = 534

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

`Z = (534 - 551)/(20)`

`Z = -0.85`

`Z = -0.85` has a pvalue of 0.1977

X = 523

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

`Z = (523 - 551)/(20)`

`Z = -1.4`

`Z = -1.4` has a pvalue of 0.0808

0.1977 - 0.0808 = 0.1169

11.69% probability that it will weigh between 523 grams and 534 grams