Question

A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on their farm follow a normal distribution with a mean of 5.85 cm and a standard deviation of 0.24 cm. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Enter your probability as a decimal value rounded to 3 decimal places.

Answer 1

`P(X>6)=P((X-\mu)/(\sigma)>(6-\mu)/(\sigma))=P(Z>(6-5.85)/(0.24))=P(z>0.625)`

And we can find this probability using the complement rule and the normal standard table or excel:

`P(z>0.625)=1-P(z<0.625)=1-0.734= 0.266`

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 diameters of mandarin oranges of a population, and for this case we know the distribution for X is given by:

`X \sim N(5.85,0.24)`

Where
`\mu=5.85` and
`\sigma=0.24`

We are interested on this probability

`P(X>6)`

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>6)=P((X-\mu)/(\sigma)>(6-\mu)/(\sigma))=P(Z>(6-5.85)/(0.24))=P(z>0.625)`

And we can find this probability using the complement rule and the normal standard table or excel:

`P(z>0.625)=1-P(z<0.625)=1-0.734= 0.266`

Answer 2

0.266

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 = 5.85, \sigma = 0.24`

Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm.

This is 1 subtracted by the pvalue of Z when X = 6.

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

`Z = (6 - 5.85)/(0.24)`

`Z = 0.625`

`Z = 0.625` has a pvalue of 0.734

1 - 0.734 = 0.266