Question

Suppose that the scores on a statewide standardized test are normally distributed with a mean of 75 and a standard deviation of 2. Estimate the percentage of scores that were (a) between 73 and 77. % (b) above 79. % (c) below 73. % (d) between 69 and 79. %

Answer 1

Solution :

Given that ,

mean = = 75

standard deviation = = 2

a) P(73 < x < 77 ) = P[(73 - 75)/ 2) < (x - ) /  < (77 - 75) /2 ) ]

= P(-1.0 < z < 1.0)

= P(z < 1.0) - P(z < -1.0)

Using z table,

= 0.8413 - 0.1587

= 0.6826

percent = 68.26%

b) P(x > 79) = 1 - p( x< 79)

=1- p P[(x - ) / < (79 - 75) / 2 ]

=1- P(z < 2.0)

= 1 - 0.9772

= 0.0228

percent = 2.28%

c) P(x < 73)

= P[(x - ) / < (73 - 75) / 2]

= P(z < -1.0)

Using z table,

= 0.1587

percent = 15.87%

d) P(69 < x < 79 ) = P[(69 - 75)/ 2) < (x - ) /  < (79 - 75) /2 ) ]

= P(-3.0 < z < 2.0)

= P(z < 2.0) - P(z < -3.0)

Using z table,

= 0.9772 - 0.0013

= 0.9759

percent = 97.59%