Assignment Five - Interpreting a Normal Distribution 1) The probability that the

Question

Assignment Five - Interpreting a Normal Distribution

1) 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

Explanation / Answer

Answer: 1) for the first part, it is a simple problem and you did it perfectly. for 50% and 60%, your P : 0.5 2.83 which you already have. now just looking at the z-table, subtract the values and you ll get 0.4977. 2) i solved this question and got the same values. so this seems right as well. 3)this is straight out of the table, all you do is look at the table and write the corresponding value which will be what you have: 0.000110 4) this is very typical exam problem and most probably this will be on your exam. but you approached it the way one should and also got the right answer. i got the same answer as well.

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