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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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(75887) (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?
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p(98 < x-bar < 102)= ?
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(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
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(b.) n = 100 ------------ (3 points)
Same procedure as "a" but let n = 100
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(c.) n = 200 ____________ (3 points)
same but n = 200
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(d.) n = 400 ____________ (3 points)
same but n = 400
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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