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vladimir2022 [97]
2 years ago
9

Number 5. I neeed the answer please and thankyou

Mathematics
2 answers:
adell [148]2 years ago
6 0

Answer:

0.4 or 2/5's of a loaf of bread

Step-by-step explanation:

For example, in number 3 the number is higher than 1 because there's more burritos than friends to share them with. In #5. the question tells you that there's 2 loaves of bread but 5 bakers meaning that each baker will have less than 1 loaf of bread. 2 divided by 5 is 2/5 or 0.4

liberstina [14]2 years ago
4 0
2/5.
2 loaves of bread is divided amongst 5 people. so 1 person gets less than a loaf of bread.
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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:

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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 = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

Z = \frac{207.9 - 208}{0.1678}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

2*0.2743 = 0.5486

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

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