The mean IQ score for 1500 students is 100, with a standard deviation of 15. Assuming the scores have a normal curve,
a. how many have an IQ between 85 and 115?
b. how many have an IQ between 70 and 130?
c. how many have an IQ over 145?
a. how many have an IQ between 85 and 115?
z(115) = (115-100)/15 = 1
z(85) = (85-100)/15 = -1
We need to consult Z-table to find P value with this Z value
P(85< x < 115) = P(-1< z < 1) = 0.6827
Quantity of people of 1500 that have IQ between 85 and 115 = 0.6827*1500 ~ 1024
b. how many have an IQ between 70 and 130?
z(115) = (130-100)/15 = 2
z(85) = (70-100)/15 = -2
P(-2< z < 2) = 0.6827
# of 1500 that have IQ between 70 and 130 = 0.6827*1500 ~ 1024
c. how many have an IQ over 145?
Z(>145)=( 145-100)/15=3
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