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vladimir2022 [97]
3 years ago
7

A random sample of 25 statistics examinations was selected. The average score in the sample was 76 with a variance of 144. Assum

ing the scores are normally distributed, the 99% confidence interval for the population average examination score is _____.
Mathematics
1 answer:
Mazyrski [523]3 years ago
6 0

Answer:

99% confidence interval for the population average examination score is between a lower limit of 69.2872 and an upper limit of 82.7128.

Step-by-step explanation:

Confidence interval = mean +/- margin of error (E)

mean = 76

variance = 144

sd = sqrt(variance) = sqrt(144) = 12

n = 25

degree of freedom (df) = n - 1 = 25 - 1 = 24

confidence level (C) = 99% = 0.99

significance level = 1 - C = 1 - 0.99 = 0.01 = 1%

t-value corresponding to 24 df and 1% significance level is 2.797

E = t×sd/√n = 2.797×12/√25 = 6.7128

Lower limit = mean - E = 76 - 6.7128 = 69.2872

Upper limit = mean + E = 76 + 6.7128 = 82.7128

99% confidence interval is (69.2872, 82.7128)

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Step-by-step explanation:

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If 4 electricians can complete a job in 8 hours how long will it take 3 electricians to complete the same job
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8 0
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A membership committee of three is formed from four eligible members. Let the eligible members be represented by A, B, C, and D.
yKpoI14uk [10]

The correct statement is:

  • There are four ways to choose the committee.
  • There are three ways to form the committee if person D must be on it.
  • If persons B and C must be on the committee, there are two ways to form the committee.

It is given that:

A membership committee of three is formed from four eligible members.

1) There are four ways to choose the committee.

This statement is true.

Since we have to choose 3 members out of the 4 members so we can use the method of combination.

2) There are three ways to form the committee if person D must be on it.

This statement is also true.

Since D has to be in the committee, this means we have to choose 2 more people out of the three people to form the committee.

3) If seven members are eligible next year, then there will be fewer combinations.

This statement is wrong.

Since we have to choose 3 members out of 7 members so the number of possible combinations will be:

There are 35 combinations possible.

4) If persons B and C must be on the committee, there are two ways to form the committee.

If B and C have to be in the committee then we have to choose just one person out of the two people left.

Hence, the statement is true.

5) If persons A and C must be on the committee, then there is only one way to form the committee.

If A and C have to be on the committee then as in the last option we have to choose any one of the two-person left.

So possible number of ways is 2.

Hence, the statement is false.

Learn more about committee here: brainly.com/question/425830

#SPJ4

<em>Your question is incomplete. Please read below to find the missing content.</em>

<em />

A membership committee of three is formed from four eligible members. Let the eligible members be represented by A, B, C, and D. The possible outcomes include S = {ABC, ABD, ACD, BCD}.

Which statements about the situation are true? Check all that apply.

There are four ways to choose the committee.

There are three ways to form the committee if person D must be on it.

If seven members are eligible next year, then there will be fewer combinations.

If persons B and C must be on the committee, there are two ways to form the committee.

If persons A and C must be on the committee, then there is only one way to form the committee.

8 0
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