Question

Question 13 of 13, Step 1 of 1 STEPHANIE HAYIBOR 9/13 Correct The mean per capita income is 20,908 dollars per annum with a standard deviation of 407 dollars per annum. What is the probability that the sample mean would differ from the true mean by more than 28 dollars if a sample of 55 persons is randomly selected? Round your answer to four decimal places. Answer How to enter your answer Tables Keypad Submit Answer

Answer 1

Solution :

Given that,

mean = $\mu$ = 20908

standard deviation = $\sigma$ = 407

n = 55

$\mu$_{$\bar x$} = $\mu$ = 20908

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 407/ $\sqrt$ 55 = 54.8799

P(20880 < $\bar x$ < 20936) = P((20880 - 20908) / 54.8799<($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (20936 - 20908) / 54.8799))

=1 - P(-0.51 < Z < 0.51)

=1 - ( P(Z < 0.51) - P(Z < -0.51) ) Using standard normal table,  

= 1 - ( 0.6950 - 0.3050 )

= 1 -0.3900

Probability = 0.6100

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