Answer:
a. 2.14% should have IQ scores between 40 and 60
b. 15.87% should have IQ scores below 80
Explanation:
* Lets explain how to solve the problem
- For the probability that a < X < b (X is between two numbers, a and b),
convert a and b into z-scores and use the table to find the area
between the two z-values.
- Lets revise how to find the z-score
- The rule the z-score is z = (x - μ)/σ , where
# x is the score
# μ is the mean
# σ is the standard deviation
* Lets solve the problem
- IQS are normally distributed with a mean of 100 and standard
deviation of 20
∴ μ = 100 and σ = 20
a.
- The IQS is between 40 and 60
∴ 40 < X < 60
∵ z = (x - μ)/σ
∴ z = (40 - 100)/20 = -60/20 = -3
∴ z = (60 - 100)/20 = -40/20 = -2
- Use the z table to find the corresponding area
∵ P(z > -3) = 0.00135
∵ P(z < -2) = 0.02275
∴ P(-3 < z < -2) = 0.02275 - 0.00135 = 0.0214
∵ P(40 < X < 60) = P(-3 < z < -2)
∴ P(40 < X < 60) = 0.0214 = 2.14%
* 2.14% should have IQ scores between 40 and 60
b.
- The IQS is below 80
∴ X < 80
∵ z = (x - μ)/σ
∴ z = (80 - 100)/20 = -20/20 = -1
- Use the z table to find the corresponding area
∵ P(z < -1) = 0.15866
∵ P(X < 80) = P(z < -1)
∴ P(X < 80) = 0.15866 = 15.87%
* 15.87% should have IQ scores below 80