Question

the mean iq score for 1500 students is 90, with a standard deviation of 15. assuming the scores will have a normal curve answer the following

How many have an IQ between 75 and 105
How many have an IQ between 60 and 120
How many have an IQ over 135

Answer 1

The question assumes normal distribution.

(a) Number of students with if between 75 and 105

The general normal distribution rule of 68% of units lie within one standard deviation from the mean can be used. However, we can calculate.

$Z=\frac{X-\mu}{\sigma}$

$P(75<X<105)=P(\frac{75-90}{15}<Z<\frac{105-90}{15})$

$=P(-1<Z<1)$

From z-table or =NORM.S.DIST(-1,TRUE) Excel formula, $P(Z<-1)=0.158655$

Similarly,

=NORM.S.DIST(1,TRUE) Excel formula gives $P(Z<1)=0.841345$

Thus, $P(-1<Z<1)=0.841345-0.158655=0.682697$

Thus, 68.2697% of the students lie between 75 and 105 $\frac{68.2697×1500}{100}=1024$

(b) between 60 and 120

The general normal distribution rule of 95% of units lie within one standard deviation from the mean can be used. However, we can calculate.

$P(60<X<120)=P(\frac{60-90}{15}<Z<\frac{120-90}{15})=P(-2<Z<2)$

From z table or =NORM.S.DIST(-2,TRUE) and=NORM.S.DIST(2,TRUE) Excel formulas, we have $P(Z<-2)=0.02275$ and $P(Z<2)=0.97725$

Thus,$P(-2<Z<2)=0.97725-0.02275=0.9544999$

Thus, 95.44999% of students lie between 60 and 120. $\frac{95.44999×1500}{100}=1431.75$ = 1432 students.

(c) over 135

$P(X>135)=1-P(Z<\frac{135-90}{15})=1-P(Z<3)$

From z table or =NORM.S.DIST(3,TRUE) Excel formula, $P(Z<3)=0.99865$

$P(X>135)=1-0.99865=0.00135$

0.135% of students have iq above 135: $\frac{0.135×1500}{100}= 2.02$

Thus, 2 students have iq above 135

Excel formulas can be used to get direct answers

(a) =(NORM.DIST(105,90,15,TRUE)-NORM.DIST(75,90,15,TRUE))×1500

(b) =(NORM.DIST(120,90,15,TRUE)-NORM.DIST(60,90,15,TRUE))×1500

(c)=(1-NORM.DIST(13,90,15,TRUE))×1500