Question

An orange juice producer buys oranges from a large orange grove that has one variety of orange. The amount of juice squeezed from these oranges is approximately normally distributed, with a mean of 5.0 ounces and a standard deviation of 0.40 ounce. Suppose that you select a sample of 25 oranges. The probability is 70% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population mean?

Answer 1

The probability is 70% that the sample mean amount of juice will be contained between 4.9168 ounces and 5.0832 ounces.

Explanation:

To solve this question, the Normal probability distribution and the Central Limit Theorem are important.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`(\sigma)/(√(n))`

In this problem, we have that:

`\mu = 5, \sigma = 0.4, n = 25, s = (0.4)/(√(25)) = 0.08`

The probability is 70% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population mean?

The lower end of this interval is the value of X when Z has a pvalue of 0.5 - 0.7/2 = 0.15

The upper end of this interval is the value of X when Z has a pvalue of 0.5 + 0.7/2 = 0.85

Lower end

X when Z has a pvalue of 0.15. So X when
`Z = -1.04`.

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

Due to the Central Limit Theorem

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

`-1.04 = (X - 5)/(0.08)`

`X - 5 = -1.04*0.08`

`X = 4.9168`

Upper end

X when Z has a pvalue of 0.15. So X when
`Z = 1.04`.

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

Due to the Central Limit Theorem

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

`1.04 = (X - 5)/(0.08)`

`X - 5 = 1.04*0.08`

`X = 5.0832`

The probability is 70% that the sample mean amount of juice will be contained between 4.9168 ounces and 5.0832 ounces.