In: Statistics and Probability
The mean capita income is 15,451 dollars per annum with a variance of 298,116.
What Is the probability that the sample mean would differ from the true mean by greater than 22 dollars if a sample of 350 persons is randomly selected? Round answer to dour decimal places.
Solution :
Given that,
mean = 
 = 15451
variance = 
2= 298116
standard deviation = 
 = 546
n = 350

= 
 = 15451

= 
 / 
n = 546 / 
350 = 29.1849
P( 15429 < 
 < 15473) = P((15429 - 15451) /29.1849 <(
- 
)
/ 
< (15473 - 15451) / 29.1849))
= P(-0.75 < Z < 0.75)
= P(Z < 0.75) - P(Z < -0.75) Using z table,
= 0.7734 - 0.2266
= 0.5468
Probability = 0.5468