Mathematics College

PLEASE HELP !! When finding the margin of error for the mean of a normally distributed population from a sample, what is the critical probability, assuming a confidence level of 58%?
0.21
0.42
0.58
0.79

Answers

Answer 1

Answer:

D) 0.79

Step-by-step explanation:

First, find alpha.

[tex]\alpha[/tex] = 1 - (confidence level / 100)

[tex]\alpha[/tex] = 1 - [tex]\frac{58}{100}[/tex]

[tex]\alpha[/tex] = 1 - 0.58

[tex]\alpha[/tex] = 0.42

Then, you can solve for the critical probability.

p* = 1 - [tex]\frac{\alpha}{2}[/tex]

p* = 1 - [tex]\frac{0.42}{2}[/tex]

p* = 1 - 0.21

p* = 0.79

Answer 2

The critical probability, assuming a confidence level of 58% will be "0.79". To understand the calculation, check below.

Probability

According to the question,

Confidence level = 58

Now,

The alpha will be:

→ α = 1 - ([tex]\frac{Confidence \ level}{100}[/tex])

By substituting the values, we get

     = 1 - ([tex]\frac{58}{100}[/tex])

     = 1 - 0.58

     = 0.42

hence,

The critical probability will be:

→ p = 1 - [tex]\frac{\alpha}{2}[/tex]

By substituting the values,

     = 1 - [tex]\frac{0.42}{2}[/tex]

     = 1 - 0.21

     = 0.79

Thus the above approach is correct.

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