Question

Suppose the scores of architects on a particular creativity tests are normally distributed with a mean of 300 and a standard deviation of 25. Using the normal curve table,

1. What percentage of architects score below 290? Hint: You need to calculate the Z-score and then find the corresponding value on the normal curve table.

2. What percentage of architects score above 330?

3. What score does an architect need on the creativity test to be in the top 10%? (Remember, after you figure out the z score, you need to convert to a raw score).

4. What is the minimum score an architect needs to be included in the bottom 25%?

Answer 1

Solution :

a.

P(x < 290) = P[(x - ) / < (290 - 300) / 25]

= P(z < -0.4) using z table

= 0.3446

percentage = 34.46%

a.

P(x > 330) = 1 - P(x < 330)

= 1 - P[(x - ) / < (330 - 300) / 25)

= 1 - P(z < 1.2) using z table   

= 1 - 0.8849

= 0.1151

percentage = 11.51%

a.

Using standard normal table,

P(Z > z) = 10%

1 - P(Z < z) = 0.1

P(Z < z) = 1 - 0.1

P(Z < 1.28) = 0.9 using z table

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 25 + 300 = 332

332 score does an architect need on the creativity test to be in the top 10%

a.

Using standard normal table,

P(Z < z) = 25%

P(Z < -0.67) = 0.25 using z table

z = -0.67

Using z-score formula,

x = z * +

x = -0.67 * 25 + 300 = 283.25

The 283.25 minimum score an architect needs to be included in the bottom 25%