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Genrish500 [490]
3 years ago
9

Because each roll has six possible even values (2, 4, 6, 8, 10, and 12) and only five possible odd values (3, 5, 7, 9, and 11),

the dice should show even values more often than odd values. Do you agree with this statement? Explain.
Mathematics
1 answer:
Natalija [7]3 years ago
8 0

lahdhd7svs7dhdhdydys

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Professor Smith has a total of 250 students,
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3/5 and 250 the number of girls is 150

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100 miles / 4 hours

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A population of plastic chairs in a factory has a weight's mean of 1.5 kg and a standard deviation of 0.1 kg . Suppose a sample
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Answer:

0.9544 = 95.44% probability that the sample mean will be within +0.02 of the population mean.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 question, we have that:

\mu = 1.5, \sigma = 0.1, n = 100, s = \frac{0.1}{\sqrt{100}} = 0.01

What is the probability that the sample mean will be within +0.02 of the population mean?

Sample mean between 1.5 - 0.02 = 1.48 kg and 1.5 + 0.02 = 1.52 kg, which is the pvalue of Z when X = 1.52 subtracted by the pvalue of Z when X = 1.48. So

X = 1.52

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

By the Central Limit Theorem

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

Z = \frac{1.52 - 1.5}{0.01}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 1.48 ​

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

Z = \frac{1.48 - 1.5}{0.01}

Z = -2

Z = -2 has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +0.02 of the population mean.

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Step-by-step explanation:

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