Question

8. Suppose you completed a study on creativity in college students and received scores for two different creativity assessments. On the first assessment, you received a 92 on a test that had a mean of 88 and a standard deviation of 3. On the second assessment, you received a 56 on a test that had a mean of 50 and a standard deviation of 8. Based on this information, you did:

a. worse on the first assessment.

b. worse on the second assessment

c. equally well on the two assessments

d. This cannot be determined without further information

9. The average pieces of candy eaten by children in your neighborhood on Halloween has a Normal distribution with a mean of 14 and a standard deviation of 5. What percentage of the children eat less than 10 pieces of candy on Halloween?

a. 80.04% b. 78.81% c. 21.19% d. 19.96%

please 8&9 and thank you

Answer 1

8.

Z score = (x - ) /

Z score first assessment = (92 - 88)/3 = 1.33

Z score second assessment = (56 - 50)/8 = 0.75

Since Z score first assessment is greater than Z score second assessment, you did

b. worse on the second assessment

9.

Percentage of the children eat less than 10 pieces of candy on Halloween

= P[x < 10]

= P[Z < (10 - 14)/5]

= P[Z < -0.8]

= 0.2119

= 21.19%

c. 21.19%