Question

In a large university, 88 percent are right-handed and 12 percent are left-handed. One hundred students are selected at random with replacement. The probability that the proportion of left-handed individuals in the sample is within 2 percentage points of the population proportion of 12 percent is closest to (use normal approximation):

Answer 1

Solution

Given that,

p = 0.12

1 - p = 1 - 0.12 = 0.88

n = 100

= p = 0.12

=  [p ( 1 - p ) / n] = [(0.12 * 0.88) / 100 ] = 0.0325

0.12 ± 0.02

=P ( 0.10 < < 0.14 )

= P[(0.10 - 0.12) / 0.0325 < ( - ) / < (0.14 - 0.12) / 0.0325 ]

= P(-0.62 < z < 0.62)

=P(z < 0.62) - P(z < -0.62 )

Using z table,   

= 0.7324 - 0.2676

= 0.4648