Mathematics, 15.04.2020 02:54 nolanreano

The weight of crackers in a box is stated to be 16 ounces. The amount that the packaging machine puts in the boxes is believed to have a Normal model with mean 16.15 ounces and standard deviation 0.3 ounces. What is the probability that the mean weight of a 50-box case of crackers is above 16 ounces?

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99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces and this can be determined by finding the z-score with the help of the given data.

Given :

The weight of crackers in a box is stated to be 16 ounces.The amount that the packaging machine puts in the boxes is believed to have a Normal model with a mean of 16.15 ounces and a standard deviation of 0.3 ounces.

The probability of the mean weight of a 50-box case of crackers is above 16 ounces can be determined by using the formula f the z-score.

$\rm z = \dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n} }}$

Now, substitute the known terms in the above formula.

$\rm z = \dfrac{16-16.15}{\dfrac{0.3}{\sqrt{50} }}$

$\rm z = \dfrac{-0.15}{0.0424}$

z = -3.54

So, z = -3.54 has a p-value of 0.0001.

P = 1 - 0.0001 = 0.9999

99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces.

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Answer from: aleilyg2005

99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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:

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

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 16.15, \sigma = 0.3, n = 50, s = \frac{0.3}{\sqrt{50}} = 0.0424$

What is the probability that the mean weight of a 50-box case of crackers is above 16 ounces?

This is 1 subtracted by the pvalue of Z when X = 16. So

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

By the Central Limit Theorem

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

$Z = \frac{16 - 16.15}{0.0424}$

Z = -3.54

Z = -3.54 has a pvalue of 0.0001

1 - 0.0001 = 0.9999

99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces

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your answer is d 2,500

step-by-step explanation:

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