Question

Assume that adults have IQ scores that are normally distributed with a mean of 95.9 and a standard deviation of 16.4. Find the following probabilities. Round your answers to three decimal places; add trailing zeros as needed.

The probability that a randomly selected adult has an IQ greater than 122.6 is .

The probability that a randomly selected adult has an IQ lower than 92.0 is .

The probability that a randomly selected adult has an IQ between 80.0 and 110.0 is .

Answer 1

Solution :

Given that ,

mean = = 95.9

standard deviation = = 16.4

1)

P(x > 122.6) = 1 - P(x < 122.6)

= 1 - P((x - ) / < (122.6 - 95.9) / 16.4)

= 1 - P(z < 1.63)

= 1 - 0.9484 Using standard normal table.

= 0.052

Probability = 0.052

2)

P(x < 92.0) = P((x - ) / < (92.0 - 95.9) / 16.4)

= P(z < -0.22)

= 0.4129 Using standard normal table,

Probability = 0.413

3)

P(80 < x < 110) = P((80 - 95.9)/ 16.4) < (x - ) /  < (110 - 95.9) / 16.4) )

= P(-0.97 < z < 0.86)

= P(z < 0.86) - P(z < -0.97)

= 0.8051 - 0.1660 Using standard normal table,  

Probability = 0.639