there are 2 classes a and b 100 students. they have the same mean 72 on a test.

Question

there are 2 classes a and b 100 students. they have the same mean 72 on a test. the standard deviation class a (6) and b (4). which class had more students scored between 60 and 80 there are 2 classes a and b 100 students. they have the same mean 72 on a test. the standard deviation class a (6) and b (4). which class had more students scored between 60 and 80 there are 2 classes a and b 100 students. they have the same mean 72 on a test. the standard deviation class a (6) and b (4). which class had more students scored between 60 and 80

Explanation / Answer

PART A.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 72
standard Deviation ( sd )= 6
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 60) = (60-72)/6
= -12/6 = -2
= P ( Z <-2) From Standard Normal Table
= 0.0228
P(X < 80) = (80-72)/6
= 8/6 = 1.3333
= P ( Z <1.3333) From Standard Normal Table
= 0.9088
P(60 < X < 80) = 0.9088-0.0228 = 0.886

PART B.
mean ( u ) = 72
standard Deviation ( sd )= 4
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 60) = (60-72)/4
= -12/4 = -3
= P ( Z <-3) From Standard Normal Table
= 0.0013
P(X < 80) = (80-72)/4
= 8/4 = 2
= P ( Z <2) From Standard Normal Table
= 0.9772
P(60 < X < 80) = 0.9772-0.0013 = 0.9759

we have more students for part B.

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