In a recent year, the scores for the reading portion of a test were normally dis

Question

In a recent year, the scores for the reading portion of a test were normally distributed, with a mean of 23.4 and a standard deviation of 5.2. Complete parts (a) through (d) below (a) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than 17. The probability of a student scoring less than 17 is (Round to four decimal places as needed.) (b) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is between 18.5 and 28.3. The probability of a student scoring between 18.5 and 28.3 is (Round to four decimal places as needed.) (c) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is more than 34.3

Explanation / Answer

given mean = 23.2

standard deviation =5.2

a) z = x^-mean ) / standard deviation

z= (17-23.2) /5.2 =-1.192

p (x<17) =p(z<-1.192) = 0.1314 from standard z table the value is found

b) z = x^-mean ) / standard deviation

z1 = (18.5-23.2) /5.2 =-0.903 from z table the value is 0.1841

z2=(28.3-23.2)/5.2 = 0.9807 from z table the value is 0.8365

p(18<x<23.2) =p(z1<x<z2) =0.8365-0.1841 =0.6524                from standard z table the value is obtained

c) z = x^-mean ) / standard deviation

       = (34.3-23.2)/5.2 =2.134

p(x>34.3)= p(1-z<2.134) = 1-0.9834=0.0166           from z table the value is obtained

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