Answer:
The percentage of people have an IQ between 52 and 148
= 0.9972 = 99.7%
Step-by-step explanation:
Step(i):-
Given mean of the normal distribution 'μ'= 100
Given standard deviation of the normal distribution 'σ' = 16
Let x₁ = 52
[tex]Z = \frac{x_{1}-mean }{S.D} = \frac{52-100}{16} = -3[/tex]
Let x₂ = 148
[tex]Z = \frac{x_{2}-mean }{S.D} = \frac{148-100}{16} = 3[/tex]
Step(ii):-
The percentage of people have an IQ between 52 and 148
P( 52≤ x≤148) = P(-3≤x≤3)
= A(3) + A(-3)
= 2 A(3) (∵A(-3) =A(3))
= 2 ×0.4986
= 0.9972
Final answer:-
The percentage of people have an IQ between 52 and 148
= 0.9972 = 99.7%
99.74% of people have an IQ between 52 and 148.
Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation[/tex]
Given that μ = 100, σ = 16.
For x = 52:
[tex]z=\frac{52-100}{16}=-3[/tex]
For x = 148:
[tex]z=\frac{148-100}{16}=3[/tex]
From the normal distribution table, P(52 < x < 148) = P(-3 < z < 3) = P(z < 3) - P(z < -3) = 0.9987 - 0.0013 = 99.74%
99.74% of people have an IQ between 52 and 148.
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