Dear Cramster: According to the ACT results from the 2004 ACT testing found that
ID: 2957964 • Letter: D
Question
Dear Cramster:According to the ACT results from the 2004 ACT testing found that students had a mean reading score of 21.3 with a standard deviation of 6.0, assuming that the scores are normally distributed.
a. Find the probability that a randomly selected student has a reading ACT score less than 20.
b. Find the probability that a randomly selected student has a reading ACT score between 18 and 24.
c. Find the probability that a randomly selected student has a reading ACT score greater than 30.
d. Find the value of the 75th percentile of ACT scores.
Thank you for this consideration.
Sincerely,
Jeannette
Explanation / Answer
Given Mean =21.3
standard deviation =6.0
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a)probability ( a randomly selected student has a reading ACT score less than 20)
=P(x<20)
=P(z<20-21.3/6)
=P(z<-1.3/6)
=P(z<-0.2167)
=0.4142
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b)probability ( a randomly selected student has a reading ACT score between 18 and 24)
=P(18<=x<=24)
=P(18-21.3/6<=z<=24-21.3/6)
=P(-3.3/6<=z<=2.7/6)
=P(-0.55<=z<=0.45)
=P(z<=0.45)+P(z<=0.55)
=0.1736+0.2088
=0.3824
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c) probability ( a randomly selected student has a reading ACT score greater than 30)
=P(x>30)
=P(z>30-21.3/6)
=P(z>8.7/6)
=P(z>1.45)
=0.5-P(z<1.45)
=0.5-0.4265
=0.0735
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d) Let z* be the z score corresponding to the 75th percentile.
Let a be the value corresponding to the 75th percentile
Then P(z<z*)=0.75
From cumulative table z*=0.67
a=0.67*6+21.3
=25.32
So the value corresponding to 75th percentile is 25.32.
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