Mathematics, 09.04.2021 02:50, cjmann71

Here is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering a question chosen at random from a universe of possible questions. (A strong student has a higher p than a weak student.) The correctness of an answer to a question is independent of the correctness of answers to other questions. Jodi is a good student for whom p = 0.81. Required:
a. Use the Normal approximation to find the probability that Jodi scores 77% or lower on a 100-question test.
b. If the test contains 250 questions, what is the probability that Jodi will score 77% or lower?
c. How many questions must the test contain in order to reduce the standard deviation of Jodi's proportion of correct answers to half its value for a 100-item test?
d. Laura is a weaker student for whom p = 0.76. Does the answer you gave in (c) for standard deviation of Jodi's score apply to Laura's standard deviation also?

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answered: mountassarhajar2008

a) 0.8461 = 84.61% probability that Jodi scores 77% or lower on a 100-question test.

b) 0.9463 = 94.63% probability that Jodi will score 77% or lower.

c) 400 questions.

d) Yes, because the formula is the same, independently of the value of p.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

p = 0.81

Question a:

100 questions means that n = 100

For the approximation, we have that:

$\mu = p = 0.81$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.81*0.19}{100}} = 0.0392$

This probability is the pvalue of Z when X = 0.77. So

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

By the Central Limit Theorem

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

$Z = \frac{0.77 - 0.81}{0.0392}$

Z = 1.02

Z = 1.02 has a pvalue of 0.8461

0.8461 = 84.61% probability that Jodi scores 77% or lower on a 100-question test.

Question b:

Now n = 250 , so:

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.81*0.19}{250}} = 0.0248$

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

$Z = \frac{0.77 - 0.81}{0.0248}$

Z = 1.61

Z = 1.61 has a pvalue of 0.9463

0.9463 = 94.63% probability that Jodi will score 77% or lower.

Question c:

The formula for the standard deviation is:

$s = \sqrt{\frac{p(1-p)}{n}}$

Meaning that it is inversely proportional to the square root of the sample size.

So, to reduce the standard deviation by half, the number of question must be multiplied by (2)^2 = 4.

100*4 = 400

So 400 questions.

d. Laura is a weaker student for whom p = 0.76. Does the answer you gave in (c) for standard deviation of Jodi's score apply to Laura's standard deviation also?

Yes, because the formula is the same, independently of the value of p.

answered: Guest

1,918

step-by-step explanation:

answered: Guest

Bi’m pretty sure hope it works