The distribution of IQ (Intelligence Quotient) is approximately normal in shape with a mean of 100 and a standard deviation of 16. According to the standard deviation rule, % of people have an IQ between 52 and 148. Do not round.

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:

Given that μ = 100, σ = 16.

For x = 52:

For x = 148:

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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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

Let x₂ = 148

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%

i think the answer is b.

step-by-step explanation:

not 100% sure though.