The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

Problems of normally distributed samples can be 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.

In this problem, we have that:

$\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12$

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when $X = 59.5$ . So

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

$Z = \frac{59.5 - 56}{1.12}$

$Z = 3.1$

$Z = 3.1$ has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0

3 years ago

Helpdisdalastquestio

Elis [28]

3 times 96 = 288
3 times 204 = 612
612-288=324
which is approximately $300

5 0

3 years ago

Read 2 more answers

Serrians triangle has a height of 3x-4 and the base length is 8x+4 what is the area of the triangle

kirill115 [55]

Area of a triangle is base × height ÷2
base=8x+4 x=-4/8 x=-1/2
height=3x-4 x=4/3

A=b×h÷2
A=(-1/2×4/3)÷2
A=(-4/6)÷2
A=(-4/6)÷2/1 to divide by fractions flip the second and multiply
A=-4/6×1/2
A=-4/12
A=-1/3

3 0

3 years ago

Help if you can pls. if you can’t see the pic don’t bother answering.

liubo4ka [24]

Ashton Martin. Because 60 divided by 20 is 3.

5 0

2 years ago