Question 515501: This is my problem and solution, please assist as my instructor says it is incorrect.
Research on new juvenile delinquents revealed that 38% of them committed another crime.
(a)What is the probability that of the last 100 new juvenile delinquents put on probation, 30 or more will commit another crime?
(b)What is the probability that 40 or fewer of the delinquents will commit another crime?
(c)What is the probability that between 30 and 40 of the delinquents will commit another crime?
This is my workings which I got incorrect
4. μ = n π = 100(0.38)
= 38
σ² = 100(0.38) (0.62)
σ² = 23.56
σ = √23.56
σ = 4.85
a. z = 38-29.5
4.85
Z = 1.753
Associated probability = 0.4599
Probability that of the last 100 new juvenile delinquents put on probation, 30 or more will commit another crime is 0.4599 + 0.5 = 0.9599
b. z = 40.5 – 38
4.85
z = 0.52
Associated probability = 0.1985
Probability that 40 or fewer of the delinquents will commit another crime is
0.5 + 0.1985 = 0.6985
c. The probability that between 30 and 40 of the delinquents will commit another crime is 0.1985 + 0.4995 = 0.698
Answer by drcole(72)   (Show Source):
You can put this solution on YOUR website! The only mistake that I see in your solution is that, in your solution to part (c), you added 0.4995 to 0.1985, instead of 0.4599.
Also, you really should make the z-score in part (a) to be z = -1.753, since 29.5 is below the mean, but you have been taught this problem differently than I would teach it.
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