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3 years ago

11

A pizza is divided into twelve equal slices. Glenn and Ben each ate 1/6 of the pizza on Monday. The next day Ben ate 1/2 of the

pizza that was leftover. How many slices of the original pizza remain? explain your reasoning.

1 answer:

Ainat [17]3 years ago

5 0

1/6 of twelve is two.

Both Glenn and Ben had two slices: 12-4=8

1/2 of eight is four.

Your final answer is four.

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0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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.

Normally distributed with a mean of 118 centimeters and a standard deviation of 8 centimeters.

This means that $\mu = 118, \sigma = 8$

Sample of 16 shells

This means that $n = 16, s = \frac{8}{\sqrt{16}} = 2$

What is the probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly?"

This is the pvalue of Z when X = 120 subtracted by the pvalue of Z when X = 116.

X = 120

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

By the Central Limit Theorem

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

$Z = \frac{120 - 118}{2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.841

X = 116

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

$Z = \frac{116 - 118}{2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.159

0.841 - 0.159 = 0.682

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

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