Assignment Five - Interpreting a Normal Distribution 1) The probability that the
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Question
Assignment Five - Interpreting a Normal Distribution1) The probability that the soft drink sample will have between 50% and 60% of the identifications correct is .4977 or 49.77%
So then P (0.5 < X < 0.6)
P (0.5 < X < 0.6) = = P (0 < Z < 2.83)
From the tables (p. 793) 0.9987 – 0.5120 = 0.4977
2) The probability is 90% that the sample percentage is contained within 5.8 % symmetrical limits of the population percentage
Find two values for K and L so that P (X < K) = 0.05and P (X > L) = 0.05.
As P (Z < -1.645) = 0.05 and P (Z > 1.645) = 0.05
Z score of K and L, is,
and
Solve for K and L
K = 0.44184 and L = 0.55816
Minimum score = 0.44184
Maximum score = 0.55816
3) The probability that the sample percentage of correct identifications is greater than 65% is 0.0000110
P (X > 0.65).
P (X > 0.65) = = P (Z > 4.24268) = 0.000110
4) More than 60% correct identifications in a sample of 200 is more likely to occur than more than 55% correct in a sample of 1000. As the sample size increases, the standard error in the denominator of the z score decreases. Then the z score becomes bigger and moves towards the tail of the curve.
P (X > 0.60) = = P (Z > 2.82845)
= 0.0023
If n = 1000, s = = = 0.015811
P (X > 0.55) = = P (Z > 3.162355 = 0.0008