Question

2. A Pew Research report indicated that 73% of teenagers aged 13-17 own smartphones. A random sample of 150 teenager is drawn.

a. Find the mean ??̅ .

b. Find the standard deviation ??̅.

c. Find the probability that more than 70% of the sampled teenagers own a smartphone.

Answer 1

Solution:

Given ,

p = 73% = 0.73(population proportion)

1 - p = 1- 0.73 = 0.27

n = 150(sample size)

Let be the sample proportion.

The sampling distribution of is approximately normal with

1)

mean = =  p = 0.73

2)

SD =   =     

=  \sqrt{0.73(1-0.73)/150}

= 0.0362491379

3)

Find P(Sample proportion is more than 70% )

= P( > 0.70 )

= P((\hat p-\mu_{\hat p})/\sigma_{\hat p} > (0.70-\mu_{\hat p})/\sigma_{\hat p} )

=  P(Z > (0.70-0.73)/0.0362491379)

= P(Z > -0.83)

= P(Z < +0.83)

= 0.7967 ...use z table

Answer: Required probability is 0.7967