Question

A study by Allstate Insurance Co. finds that 82% of teenagers have used cell phones while driving (the Wall Street Journal, May 5, 2010). Suppose a random sample of 100 teen drivers is taken. What is the probability that the sample proportion is within ± 0.02 of the population proportion?

Answer 1

The probability that the sample proportion is within ± 0.02 of the population proportion is 0.3328

How to determine the probability?

The given parameters are:

• Sample size, n = 100
• Population proportion, p = 82%

Start by calculating the mean:

$\mu = np$

$\mu = 100 * 82\%$

$\mu = 82$

Calculate the standard deviation:

$\sigma = \sqrt{\mu(1 - p)$

$\sigma = \sqrt{82 * (1 - 82\%)$

$\sigma = 3.84$

Within ± 0.02 of the population proportion are:

$x_{min} = 82 * (1 - 0.02) = 80.38$

$x_{max} = 82 * (1 + 0.02) = 83.64$

Calculate the z-scores at these points using:

$z = \frac{x - \mu}{\sigma}$

So, we have:

$z_1 = \frac{80.36 - 82}{3.84} = -0.43$

$z_2 = \frac{83.64 - 82}{3.84} = 0.43$

The probability is then represented as:

P(x ± 0.02) = P(-0.43 < z < 0.43)

Using the z table of probabilities, we have:

P(x ± 0.02) = 0.3328

Hence, the probability that the sample proportion is within ± 0.02 of the population proportion is 0.3328

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