SOLUTION: Please help I'm struggling to solve this Z-SCORE Suppose you have breathalyzer reading scores for 500 adults recently arrested for DUI. In the initial stages of your data ana

Question 1040314: Please help I'm struggling to solve this
Z-SCORE
Suppose you have breathalyzer reading scores for 500 adults recently arrested for DUI. In the initial stages of your data analysis, you determine that the readings for the sampling distribution of scores is approximately normally distributed with a mean reading of μ = 0.071 and a standard deviation of σ = 0.011. During pre-trial, the pros�ecutor asks you to find out how one of the arrestees, who had a reading of 0.073 (X), compares to the others arrested. Use these parameters to answer the following questions:
a.What proportion and how many of the readings were above .083 (X> 0.083)? 0.073>0.083
b.What proportion and how many of the readings were above .035 (X> 0.035)?
***For this question: -Draw the standard normal curve for both (a) and (b) [separately], and take a picture of each. Make sure they show the areas properly including the Z scores.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Suppose you have breathalyzer reading scores for 500 adults recently arrested for DUI. In the initial stages of your data analysis, you determine that the readings for the sampling distribution of scores is approximately normally distributed with a mean reading of μ = 0.071 and a standard deviation of σ = 0.011.
During pre-trial, the prosecutor asks you to find out how one of the arrestees, who had a reading of 0.073 (X), compares to the others arrested. Use these parameters to answer the following questions:
a.What proportion and how many of the readings were above .083 (X> 0.083)?
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z(0.083) = (0.083-0.071)/0.011 = 1.09
P(X < 0.083) = P(z > 1.09) = normalcdf(1.09,100) = 0.1377
# of readings above = 0.1377*500 = 69 when rounded up
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b.What proportion and how many of the readings were above .035 (X> 0.035)?
z(0.035) = (0.035-0.071)/0.011 = -3.27
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P(X > 0.035) = P(z > -3.27) = normalcdf(-3.27,100) = 0.9995
# of readings above = 499 when rounded down
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Cheers,
Stan H.
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