Question

The number of words in the active vocabulary of children of a particular age is normally distributed with a mean of 3000 and a standard deviation of 500.

Use the Normal Curve Table to determine the number of words a child needs to have in his/her vocabulary in order to be included

(a) among the top 5% of vocabulary scores

(b) among the top 20% of vocabulary scores

(c) among the bottom 10% of vocabulary scores

Answer 1

Solution :

mean = = 3000

standard deviation = = 500

Using standard normal table,

a ) P(Z > z) = 5%

1 - P(Z < z) = 0.05

P(Z < z) = 1 - 0.05 = 0.95

P(Z < 1.645 ) = 0.95

z = 1.65

Using z-score formula,

x = z * +

x = 1.65 * 500 + 3000

= 3825

Vocabulary scores = 3825

b ) P(Z > z) = 20%

1 - P(Z < z) = 0.20

P(Z < z) = 1 - 0.20 = 0.80

P(Z < 0.8416 ) = 0.80

z = 0.84

Using z-score formula,

x = z * +

x = 0.84 * 500 + 3000

= 3425

Vocabulary scores = 3425

c ) P(Z < z) = 10%

P(Z < z) = 0.10

P(Z < -1.282 ) = 0.10

z = -1.28

Using z-score formula,

x = z * +

x = -1.28 * 500 + 3000

= 2360

Vocabulary scores = 2360