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

Answer:

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 $\sigma = \sqrt{0.25} = 0.5$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3.75 - 4}{0.5}$

$Z = -0.5$

$Z = -0.5$ 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.

Answer 2

Answer:

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Step-by-step explanation: