Question

The weight of a product is normally distributed with a mean of four ounces and a variance of .25 squared ounces. What is the probability that a randomly selected unit from a recently manufactured batch weighs more than 3.75 ounces? Do not round intermediate calculations. Round your final answer to four decimals.

Answer 1

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Explanation:

Answer 2

69.15% probability that a randomly selected unit from a recently manufactured batch weighs more than 3.75 ounces.

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:

Mean of four ounces, so
`\mu = 4`

Variance of .25 squared ounces. The standard deviation is the square root of the variance, so
`\sigma = √(0.25) = 0.5`

What is the probability that a randomly selected unit from a recently manufactured batch weighs more than 3.75 ounces?

This is 1 subtracted by the pvalue of Z when X = 3.75. So

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

`Z = (3.75 - 4)/(0.5)`

`Z = -0.5`

`Z = -0.5` has a pvalue of 0.3085.

So there is a 1-0.3085 = 0.6915 = 69.15% probability that a randomly selected unit from a recently manufactured batch weighs more than 3.75 ounces.