One subject is tested resulting in a perception test score of 1800. Find her sta
ID: 3321910 • Letter: O
Question
One subject is tested resulting in a perception test score of 1800. Find her standard/z-score. What does this z-score tell you? If one subject is randomly selected and tested, find the probability of the score being higher than 1800. With regard to this distribution in (a), determine P(x 1500). Is this is likely event? Interpret. A job requirement stipulates that all applicants for employment must score in the top 80% of this vision test. What test score separates the top 80% of all scores from the bottom 20%? If 50 subjects are randomly selected and tested, describe (or sketch and label) the sampling distribution of the means for this sample size as predicted by the CLT. If one sample of 50 subjects is randomly selected, find the probability that the sample mean will be between 2000 and 2050. Based on this value, is this event likely or unlikely to occurr? If one sample of 50 subjects is randomly selected, find the probability that the sample mean will be over 2050.Explanation / Answer
Answer to the question is as follows:
1.
Pop.Mean = 2020
Pop.Stdev = 250
Sample.Mean = 1800
Zscore = (Smaple mean - Pop.mean)/Stdev = (1800-2020)/250 = -0.88
2.
P(X>1800)
= P(Z>-.88)
= .8106
3.
P(X<1500)
= P(Z< (1500-2020) / 250)
= P(Z<-2.08)
= .0188
4.
For this first take out the Z for bottom 20% threshold. Z = -.84
Therefore the score will be X+Z*Sigma = 2020 - .84*250 = 1810
5.
n =50
Then the sample mean becomes same as population mean = 2020
Then the sample deviation becomes population deviation divided by sqrt(50) = 250/sqrt(50) = 35.355
6.
P(2000<X<2050)
= P( (2000-2020)/(250/sqrt(50)) <Z< (2050-2020)/(250/sqrt(50))
= P(-0.566<Z<.849)
= .8019-.2859
= .516
7.
P(X>2050)
= P(Z> (2050-2020)/(250/sqrt(50)))
= P(Z>.849)
= 1-.8019
= .1981
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