Question

17. Assume that adults have IQ scores that are normally distributed with a mean of 102.5 and a standard deviation of 16.5. Find the probability that a randomly selected adult has an IQ greater than 128.8. ​(Hint: Draw a​ graph.) The probability that a randomly selected adult from this group has an IQ greater than 128.8 is. ​ (Round to four decimal places as​ needed.)

35. Use a normal approximation to find the probability of the indicated number of voters. In this​ case, assume that 199 eligible voters aged​ 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged​ 18-24, 22% of them voted.

Probability that fewer than 49 voted

The probability that fewer than 49 of 199 eligible voters voted is

36. Use a normal approximation to find the probability of the indicated number of voters. In this​ case, assume that 132 eligible voters aged​ 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged​ 18-24, 22% of them voted.

Probability that exactly 31 voted

The probability that exactly than 31 of 132 eligible voters voted is

Answer 1

Question 17

X ~ N ( µ = 102.5 , σ = 16.5 )
P ( X > 128.8 ) = 1 - P ( X < 128.8 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 128.8 - 102.5 ) / 16.5
Z = 1.5939
P ( ( X - µ ) / σ ) > ( 128.8 - 102.5 ) / 16.5 )
P ( Z > 1.5939 )
P ( X > 128.8 ) = 1 - P ( Z < 1.5939 )
P ( X > 128.8 ) = 1 - 0.9445
P ( X > 128.8 ) = 0.0555

Question 35

Using Normal Approximation to Binomial
Mean = n * P = ( 199 * 0.22 ) = 43.78
Variance = n * P * Q = ( 199 * 0.22 * 0.78 ) = 34.1484
Standard deviation = √(variance) = √(34.1484) = 5.8437

Condition check for Normal Approximation to Binomial
n * P >= 10 = 199 * 0.22 = 43.78
n * (1 - P ) >= 10 = 199 * ( 1 - 0.22 ) = 155.22

P ( X < 49 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 49 - 0.5 ) = P ( X < 48.5 )

X ~ N ( µ = 43.78 , σ = 5.8437 )
P ( X < 48.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 48.5 - 43.78 ) / 5.8437
Z = 0.81
P ( ( X - µ ) / σ ) < ( 48.5 - 43.78 ) / 5.8437 )
P ( X < 48.5 ) = P ( Z < 0.81 )
P ( X < 48.5 ) = 0.791

Question 36

Using Normal Approximation to Binomial
Mean = n * P = ( 132 * 0.22 ) = 29.04
Variance = n * P * Q = ( 132 * 0.22 * 0.78 ) = 22.6512
Standard deviation = √(variance) = √(22.6512) = 4.7593

Condition check for Normal Approximation to Binomial
n * P >= 10 = 132 * 0.22 = 29.04
n * (1 - P ) >= 10 = 132 * ( 1 - 0.22 ) = 102.96

P ( X = 31 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 31 - 0.5 < X < 31 + 0.5 ) = P ( 30.5 < X < 31.5 )

X ~ N ( µ = 29.04 , σ = 4.7593 )
P ( 30.5 < X < 31.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 30.5 - 29.04 ) / 4.7593
Z = 0.31
Z = ( 31.5 - 29.04 ) / 4.7593
Z = 0.52
P ( 0.31 < Z < 0.52 )
P ( 30.5 < X < 31.5 ) = P ( Z < 0.52 ) - P ( Z < 0.31 )
P ( 30.5 < X < 31.5 ) = 0.6985 - 0.6217
P ( 30.5 < X < 31.5 ) = 0.0767