Download Part II – Free Response 1a) Mean = 0.8 Standard deviation

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Part II – Free Response
1a) Mean = 0.8
Standard deviation = √
(0.8)(0.2)
120
= 0.0365
1b) np = (0.8)(120) = 96
n(1 – p) = (0.2)(120) = 24
*Both are more than 10 so the sampling distribution is normal
1c) Shape – Not normal
np = (0.8)(30) = 24
n(1 – p) = (0.2)(30) = 6
*Since n(1 – p) is less than 10, then the sampling distribution is not normal
Center - Same
Mean = 0.8
Spread – Larger
Standard deviation = √
(0.8)(0.2)
30
= 0.073
1d) The largest possible sample that could be drawn is 50 because that is 10% of the population size (500).
2a) Mean = 0.22
Standard deviation = √
2b) z =
0.3−0.22
0.0586
(0.22)(0.78)
50
= 0.0586
= 1.36
P(z > 1.36) = normalcdf(1.36, e99,0,1) = 0.0869
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