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MakcuM [25]
3 years ago
8

Dorothy Little purchased a mailing list of 2,000 names and addresses for her mail order business, but after scanning the list sh

e doubts the authenticity of the list. She randomly selects five names from the list for validation. If 40% of the names on the list are non-authentic, and x is the number on non-authentic names in her sample, the expected (average) value of x is ______________.
a) 2.50
b) 2.00
c) 1.50
d) 1.25
e) 1.35
Mathematics
1 answer:
DaniilM [7]3 years ago
8 0

Answer:

b) 2.00

Step-by-step explanation:

For each name, there are only two possible outcomes. Either it is authentic, or it is not. The probability of a name being authentic is independent of any other name. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

E(X) = np

She randomly selects five names from the list for validation.

This means that n = 5

40% of the names on the list are non-authentic

This means that p = 0.4

The expected (average) value of x is

E(X) = np = 5*0.4 = 2

The correct answer is given by option B.

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26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics
hjlf

The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.

Then they are given a post-test once the professor has concluded lecturing on the material.

Pre-test and post-test scores for 4 students in an elementary statistics class are given below.

Then we have

\mu _d = \mu _{post} - \mu _{pre}

Then the null hypotheses and alternative hypotheses will be

H₀: \mu _d = 0

Hₐ: \mu _d > 0

Then the test statistic will be

\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5

Then

\rm S_d = 6.02

The test statistic value is given by

\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16

Since this is a right-tailed test, so the critical value is given by

\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353

Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

Hence, we can conclude that \mu _d > 0 that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

brainly.com/question/9504281

#SPJ1

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