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The Internal Revenue Service estimates that 8% of all taxpayers filling out forms make mistakes. An IRS employee announces at lunch one day that she is going to check 25 randomly selected forms after lunch. Calculate the mean and standard deviation of the number of forms with mistakes she should expect to find.

Sagot :

Answer:

Mean = 2

Standard deviation = 1.356

Step-by-step explanation:

We are given;

Probability of taxpayers that make mistakes when filling out forms: p = 0.08

Sample size; n = 25

In binomial distribution;

Means is;

μ = np

Thus;

μ = 25 × 0.08

μ = 2

Formula for standard deviation here is;

σ = √(np(1 - p))

σ = √(25 × 0.08(1 - 0.08))

σ = 1.356

The required mean is 2 and standard deviation is 1.356 of the number of forms with mistakes she should expect to find.

Given that,

The Internal Revenue Service estimates that 8% of all taxpayers filling out forms make mistakes.

An IRS employee announces at lunch one day that she is going to check 25 randomly selected forms after lunch.

We have to determine,

The mean and standard deviation of the number of forms with mistakes she should expect to find.

According to the question,

Probability of taxpayers that make mistakes when filling out forms: p = 8%= 0.08

And sample size; n = 25

Therefore,

In binomial distribution mean is determine by the formula,

[tex]\mu = np\\\\\mu = 25 \times 0.08\\\\\mu = 2[/tex]

The mean of the number of forms with mistake she should expect to find is 2.

And, Standard deviation is determined by the formula,

[tex]\sigma = \sqrt{np(1-p)}\\\\\sigma = \sqrt{25\times 0.08 (1-0.08)}\\\\\sigma = \sqrt{2\times0.92}\\\\\sigma =\sqrt{ 1.84}\\\\\sigma = 1.356[/tex]

The standard deviation of the number of forms with mistake she should expect to find is 1.356.

Hence, The required mean is 2 and standard deviation is 1.356 of the number of forms with mistakes she should expect to find.

To know more about Standard deviation click the link given below.

https://brainly.com/question/19793958

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