When the population mean and standard deviation are known, what distribution is used for the analysis of the sample?.

Mathematics

1 answer:

Kitty [74]2 years ago

7 0

When the population mean and standard deviation are known, you use the standard normal distribution

<h3>How to determine the distribution?</h3>

There are several probability distributions; these include

Normal distribution
Poisson distribution
Chi square distribution
Binomial distribution
Etc

Of all these distribution, only the standard normal distribution can be used when the population mean and standard deviation are known,

Note that it is also referred to as the z-distribution

Read more about probability distributions at:

brainly.com/question/24756209

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Answer:

a) By the Central Limit Theorem, the distribution would be approximately normal, with mean $\mu = 0.45$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497$

b) 4.02 standard deviations below the mean.

c) Making 25 or less shots has a pvalue of -4.02. Z-scores lower than -2 are considered surprising, so yes, it would be surprising for him to make only 25 of these shots.

Step-by-step explanation:

To solve this question, we are going to 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.

For proportions, we have that the mean is $\mu = p$ and the standard deviation is $s = \sqrt{\frac{p(1-p)}{n}}$

In this problem, we have that:

$p = 0.45, n = 100$

a)By the Central Limit Theorem, the distribution would be approximately normal, with mean $\mu = 0.45$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497$

b)This is Z when X = 0.25.

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

By the Central Limit Theorem

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