A random sample of size n = 241 is taken from a population of size N = 5,588 with mean μ = −68 and variance σ^{2} = 183. [You may find it useful to reference the z table.]
a-1. Is it necessary to apply the finite population correction factor?
Yes
No
a-2. Calculate the expected value and the standard
error of the sample mean. (Negative values should be
indicated by a minus sign. Round "standard error" to 2
decimal places.)
Expected value | |
Standard error |
b. What is the probability that the sample mean is
between −70 and −66? (Round “z” value to 2 decimal
places, and final answer to 4 decimal places.)
c. What is the probability that the sample mean is
greater than −67? (Round “z” value to 2 decimal places, and
final answer to 4 decimal places.)
a1)
yes
a2)
Expected value = -68
fpc = sqrt ((N - n) / (N - 1))
= sqrt((5588 - 241)/(5588-1))
= 0.9783
variance = 183
std.dev =sqrt(183) = 13.5277
std.error = (s/sqrt(n)) *fpc
= (13.5277/sqrt(241)) * 0.9783
= 0.85
b)
P(-70 < x < -66)
= P((-70 - (-68))/0.85 < z < ( -66 - (-68))/0.85)
= P(-2.3529 < z < 2.3529)
= P(z < 2.3529) - P(z< -2.3529)
= 0.9907 - 0.0093
= 0.9814
c)
P(x> -67)
= P(z > (-67 - (-68))/0.85)
= P(z > 1.1765)
= 1- P(z< 1.1765)
= 1 - 0.8803
= 0.1197
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