In a recent year scores on a standardized test for high school students with a 3
ID: 3396023 • Letter: I
Question
In a recent year scores on a standardized test for high school students with a 3.50 to 4.00 grade point average were normally distributed with a mean of 37.8 and a standard deviation of 1.5. A student with a 3.50 to 4.00 grade point average who took the standardized test is randomly selected
The probability of the student scoring less than 36 is Find the probabiltiy that the students test score is between 35.2 and 40.4 the probability of student scoreing between 35.2 and 40 is ?
find the probability that the students score is more than 39.1 the probability of a student scoring more than 39.1 is?
Explanation / Answer
let X be the random variable denoting the scores for high school students with a 3.50 to 4.00 grade point average.
in the question it is given that X is normally distributed with mean 37.8 and standard deviation 1.5
so X~N(37.8,1.52)
so the probability that student scoring less than 36 is
P[X<36]=P(X-37.8)/1.5<(36-37.8)/1.5]=P[Z<-1.2]=0.115070 [answer] [using minitab] [Z~N(0,1)]
the probability that the student's test score is between 35.2 and 40.4 is
P[35.2<X<40.4]=P[(35.2-37.8)/1.5<(X-37.8)/1.5<(40.4-37.8)/1.5]=P[-1.73<Z<1.73]=P[Z<1.73]-P[Z<-1.73]
=0.958185-0.041815=0.916370 [answer] [using minitab]
the probability that the student's test score is more than 39.1 is
P[X>39.1]=P(X-37.8)/1.5>(39.1-37.8)/1.5]=P[Z>0.8667]=1-P[Z<0.8667]=1-0.806947=0.193053 [answer] [using minitab]
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