Question

Q15: (1 mark) An article in Parenting magazine reported that 60% of Americans needed a vacation after visiting their families for the holidays. Suppose this is the true proportion of Americans who feel this way. A random sample of 100 Americans is taken. What is the probability that less than 50% of the people in the sample feel that they need a vacation after visiting their families for the holidays?

1. 0.4000.

2. 0.1446.

3. 0.0207.

4. 0.0062.

Answer 1

Solution

Given that,

p = 0.60

1 - p = 1-0.60=0.40

n = 100

= p =0.60

=  [p ( 1 - p ) / n] =   [(0.60*0.40) / 100 ] = 0.04899

P( <0.50 ) =

= P[( - ) / < (0.50-0.60) /0.04899 ]

= P(z <-2.04 )

Using z table,   

=0.0207

probability=0.0207