Question

1. The average score and standard deviation of a kinesiology test were; Mean: 80 Std Dev: 4
   1. What percentage of scores on the test were below 76? Show your calculations. 3 points
   2. What percentage of scores on the test were above 83? Show your calculations. 3 points
   3. What percentage of scores on the test were between 76 and 83? Show your calculations. 1 point
   4. What score on the test did a student earn if that student was at the 25^th percentile on that test? Show your calculations. 3 points

Answer 1

Solution :

Given that ,

mean = = 80

standard deviation = = 4

a) P(x < 76) = P[(x - ) / < (76 - 80) / 4]

= P(z < -1)

Using z table,

= 0.1587

b) P(x > 83) = 1 - p( x< 83)

=1- p P[(x - ) / < (83 - 80) / 4]

=1- P(z < 0.75)

Using z table,

= 1 - 0.7734

= 0.2266

c) P(76 < x < 83) = P[(76 - 80)/ 4) < (x - ) /  < (83 - 80) /4 ) ]

= P(-1 < z < 0.75)

= P(z < 0.75) - P(z < -1)

Using z table,

= 0.7734 - 0.1587

= 0.6147

d) Using standard normal table,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

x = z * +

x = -0.6745 * 4 + 80

x = 77.30