Question

It is known that 70% of the students in a summer course are distracted in class observing details of the 2018 World Cup in Russia. If we take a sample of 50 students, what is the probability that more than 60% of the students in the sample get distracted by watching details of the World Cup during classes? (using the approximation to Normal). Answer=.9385

Answer 1

Solution

Given that,

p = 0.70

1 - p = 1 - 0.70 = 0.30

n = 50

= p = 0.70

=  [p( 1 - p ) / n] = [(0.70 * 0.30) / 50 ] = 0.064807407

P( > 0.60) = 1 - P( < 0.60)

= 1 - P(( - ) / < (0.60 - 0.70) / 0.0648 )

= 1 - P(z < 1.543)

Using z table

= 1 - 0.0615

= 0.9385