Question

1. A research report indicated that in 2018, 59% of all 6^th graders owned a cell phone. A random sample of 151 6^th graders is drawn.

1. Is it appropriate to use the normal approximation for this situation? Why or why not?

2. Find the probability that the sample proportion of 6^th graders who own a cell phone is more than 64.5%. (Find P(p > 0.645)).

3. Calculate the probability that the sample proportion of 6^th graders who own a cell phone is between 54% and 66%. (Find P(.54 < p < 0.66)).

Answer 1

Solution

Given that,

p = 0.59

1 - p = 1 - 0.59 = 0.41

n = 151

a) np 10 and n(1 - p) 10

151 * 0.59 = 89.09 10 and 151 * 0.41 = 61.91 10

The sampling distribution is approximately normal,

= p = 0.59

=  [p( 1 - p ) / n] = [(0.59 * 0.41) / 151 ] = 0.040

b) P( > 0.645) = 1 - P( < 0.645 )

= 1 - P(( - ) / < (0.645 - 0.59) / 0.040)

= 1 - P(z < 1.375)

Using z table

= 1 - 0.9154

= 0.0846

c) P( 0.54 < < 0.66 )

= P[(0.54 - 0.59) / 0.040 < ( - ) / < (0.66 - 0.59) / 0.040 ]

= P(-1.25 < z < 1.75)

= P(z < 1.75) - P(z < -1.25)

Using z table,   

= 0.9599 - 0.1056

= 0.8543