Question

An interpersonally Reactivity Scale was give to a POPULATION of juvenile offenders to measure their levels of empathy for victims. Assume scores are NORMALLY DISTRIBUTED with a MEAN = 7.66 and a STANDARD DEVIATION = 1.75.

a. What is the probability of someone scoring a 9.02 of higher?

b. 50% of offenders clustered around the mean would score between what and what?

c. 95% of the offenders clustered around the mean would score between what and what?

Answer 1

Solution :

Given that,

mean = = 7.66

standard deviation = =27

a ) P (x > 9.02 )

= 1 - P (x < 9.02 )

= 1 - P ( x -  / ) < ( 9.02 - 7.66 /1.75.)

= 1 - P ( z < 1.36 / 1.75. )

= 1 - P ( z < 0.78 )

Using z table

= 1 - 0.7823

= 0.2177

Probability = 0.2177

b ) P(-z < Z < z) = 50%
P(Z < z) - P(Z < z) = 0.50
2P(Z < z) - 1 = 0.50
2P(Z < z ) = 1 + 0.50
2P(Z < z) = 1.50
P(Z < z) = 1.50 / 2
P(Z < z) = 0.75
z = 0.67 znd z = - 0.67

Using z-score formula,

x = z * +

x = 0.67 * 1.75. + 7.66

= 8.83

x = 8.83

x = z * +

x = - 0.67 * 1.75. + 7.66

= 6.49

x = 6.49

c ) P(-z < Z < z) = 95%
P(Z < z) - P(Z < z) = 0.95
2P(Z < z) - 1 = 0.95
2P(Z < z ) = 1 + 0.95
2P(Z < z) = 1.95
P(Z < z) = 1.95 / 2
P(Z < z) = 0.975
z = 1.96 z = - 1.96

Using z-score formula,

x = z * +

x = 1.96 * 1.75. + 7.66

= 11.09

x = 11.09

x = z * +

x = - 1.96 * 1.75. + 7.66

= 4.23

x = 4.23