# SOLUTION: Assume we draw a simple random sample from a population having a mean of 100 and a standard deviation of 16. What is the probability that a sample mean will be within plus-or-minus

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 Click here to see ALL problems on Probability-and-statistics Question 244552: Assume we draw a simple random sample from a population having a mean of 100 and a standard deviation of 16. What is the probability that a sample mean will be within plus-or-minus two of the population mean for each of the following sample sizes? (a.) n = 50 ____________ (3 points) (b.) n = 100 96.8 to 103.2 (3 points) (c.) n = 200 ____________ (3 points) (d.) n = 400 ____________ (3 points) Found 2 solutions by stanbon, desire150:Answer by stanbon(57282)   (Show Source): You can put this solution on YOUR website!Assume we draw a simple random sample from a population having a mean of 100 and a standard deviation of 16. What is the probability that a sample mean will be within plus-or-minus two of the population mean for each of the following sample sizes? --- p(98 < x-bar < 102)= ? --- (a.) n = 50 ____________ (3 points) t(98) = (98-100)/[16/sqrt(50)] = -0.884 t(102) = (102-100)/[16/sqrt(50)] = +0.884 P(98 < x-bar < 102) = P(-0.884 < t < 0.884 when df = 49) = 0.619 ====================================================================== (b.) n = 100 ------------ (3 points) Same procedure as "a" but let n = 100 ------------- (c.) n = 200 ____________ (3 points) same but n = 200 ------------------- (d.) n = 400 ____________ (3 points) same but n = 400 ======================================== Cheers, Stan H. Answer by desire150(1)   (Show Source): You can put this solution on YOUR website!Here is the answers with the solutions (b.) n = 100 ____________ (3 points) t(98) = (98-100)/[16/sqrt(100)] = -1.25 t(102) = (102-100)/[16/sqrt(100)] = +1.25 P(98 < x-bar < 102) = P(-1.25 < t < 1.25 when df = 49) = c.) n = 200 ____________ (3 points) t(98) = (98-100)/[16/sqrt(200)] = -1.76 t(102) = (102-100)/[16/sqrt(200)] = +1.76 P(98 < x-bar < 102) = P(-1.76 < t < 1.76 when df = 49) = d.) n = 400 ____________ (3 points) t(98) = (98-100)/[16/sqrt(400)] = -2.5 t(102) = (102-100)/[16/sqrt(400)] = +2.5 P(98 < x-bar < 102) = P(-2.5< t < 2.5 when df = 49) = How do I solve for P