Question

1. A scale measuring prejudice has been administered to a large sample of respondents. The distribution of scores is approximately normal, with a mean of 30 and a standard deviation of 5. What percent of the sample had scores…a. Below 20? b. Below 40? c. Between 30 and 40? d. Between 35 and 45? e. Above 25? f. Above 35?

Answer 1

Solution :

Given that ,

mean = = 30

standard deviation = = 5

a.

P(x < 20) = P[(x - ) / < (20 - 30) / 5]

= P(z < -2)

= 0.0228

Percent = 2.28%

b.

P(x < 40) = P[(x - ) / < (40 - 30) /5 ]

= P(z < 2)

= 0.9772

Percent = 97.72%

c.

P(30 < x < 40) = P[(30 - 30)/ 5) < (x - ) /  < (40 - 30) / 5) ]

= P(0 < z < 2)

= P(z < 2) - P(z < 0)

= 0.9772 - 0.5

= 0.4772

Percent = 47.72%

d.

P(35 < x < 45) = P[(35 - 30)/ 5) < (x - ) /  < (45 - 30) / 5) ]

= P(1 < z < 3)

= P(z < 3) - P(z < 1)

= 0.9987 - 0.8413

= 0.1574

Percent = 15.74%

e.

P(x > 25) = 1 - P(x < 25)

= 1 - P[(x - ) / < (25 - 30) / 5]

= 1 - P(z < -1)

= 1 - 0.1587

= 0.8413

Percent = 84.13%

f.

P(x > 35) = 1 - P(x < 35)

= 1 - P[(x - ) / < (35 - 30) / 5]

= 1 - P(z < 1)

= 1 - 0.8413

= 0.1587

Percent = 15.87%