Mathematics, 05.02.2021 22:10 kayranicole1

An experiment on the teaching of reading compares two methods, A and B. The response variable is the Degree of Reading Power (DRP) score. The experimenter uses Method A in a class of 100 students and Method B in a comparable class of 100 students. Students are assigned to the two classes at random. Suppose that in the population of all children of this age the DRP score has the N(75, 30) distribution if Method A is used and the N(74, 40) distribution if Method B is used. Let us call the mean DRP score for 100 students in the A group. Let us call the mean DRP score for 100 students in the B group ý. Required:
What is the probability that the mean score for the B group will be at least five points higher than the mean score for the A group?

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0% probability that the mean score for the B group will be at least five points higher than the mean score for the A group

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution, the central limit theorem, and subtraction of normal variables:

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.

Subtraction of normal variables:

When we subtract two normal variables, the mean is the subtraction of the mean of each variable, while the standard deviation is the square root of the sum of each variance.

Method A:

One student: N(75,30), so $\mu = 75, \sigma = 30$

Samples of 100 students: $n = 100, s = \frac{30}{\sqrt{100}} = 0.3$

Method B:

One student: N(74,40), so $\mu = 74, \sigma = 40$

Samples of 100 students: $n = 100, s = \frac{40}{\sqrt{100}} = 0.4$

What is the probability that the mean score for the B group will be at least five points higher than the mean score for the A group?

This is the probability that B - A >= 5. So

$\mu_{B-A} = \mu_B - \mu_A = 74 - 75 = -1$

$s_{B-A} = \sqrt{s_A^{2}+S_B^{2}} = \sqrt{(0.3)^2+(0.4)^2} = 0.5$

We have to find 1 subtracted by the pvalue of Z when X = 5. So

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

By the Central Limit Theorem

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

$Z = \frac{5 - (-1)}{0.5}$

Z = 12

Z = 12 has a pvalue of 1

1 - 1 = 0, so

0% probability that the mean score for the B group will be at least five points higher than the mean score for the A group

Answer from: Quest

It’s either a or c as the answer.

Answer from: Quest

hello! just here for points! though i wish i knew!

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