Answer:
Step-by-step explanation:
Let X be the IQ
IQ is normally distributed with a mean of 100 and a standard deviation of 15.
a) the probability that this person has an IQ greater than 95
=[tex]P(X\geq 95)\\=P(Z\geq \frac{95-100}{15} \\=P(Z\geq -0.33)\\=0.1293+0.5\\=0.6293[/tex]
=62.93%
b) the probability that this person has an IQ less than 125
[tex]=P(X\leq 125) =P(X\leq 1.67)\\=0.5-0.4525\\=0.0475[/tex]
=4.75%
c) Sample size =500
[tex]P(X<110)=P(Z<0.67)\\=0.2486[/tex]
No of people = 0.2486(500) =124.3
d) [tex]P(x>140)\\= P(Z>2.6)\\=0.5-0.4953\\=0.0047[/tex]
No of persons = 0.0047(500) = 2.35