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A headline in a certain newspaper states that​ "most stay at first job less than 2​ years." That headline is based on an online poll of 400 college graduates. Among those​ polled, 78 % stayed at their first​ full-time job less than 2 years. Complete parts​ (a) through​ (d). a. Assuming that​ 50% is the true percentage of graduates who stay at their first job less than two​ years, find the mean and standard deviation of the numbers of such graduates in randomly selected groups of 400 graduates.

Respuesta :

Answer:

Mean, μ = 200

Standard deviation, σ = 10

Step-by-step explanation:

Data provided:

a) true percentage of graduates who stay at their first job less than two​ years

p = 50% = 0.5

q = 1 - p = 1 - 0.5 = 0.5

Number of graduates selected in a group, n = 400

Now,

The mean is calculated as:

Mean, μ = p × n

or

μ = 0.5 × 400 = 200

and,

standard deviation, σ = [tex]\sqrt{npq}[/tex]

or

σ = [tex]\sqrt{400\times0.5\times0.5}[/tex]

or

σ = √100

or

σ = 10

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