Question

if 15% of adults in a certain country work from home, what is the probability that fewer than 60 out of a random sample of 500 adults will work from home? (round your final answer to 4 decimal places.)

Explanation:

Step1: Calculate the mean and standard deviation

The mean of a binomial distribution is \(\mu = np\), where \(n = 500\) and \(p=0.15\). So \(\mu=500\times0.15 = 75\).
The standard deviation is \(\sigma=\sqrt{np(1 - p)}\), so \(\sigma=\sqrt{500\times0.15\times(1 - 0.15)}=\sqrt{500\times0.15\times0.85}=\sqrt{63.75}\approx7.984\).

Step2: Use the continuity correction

We want \(P(X < 60)\). Using the continuity correction for the normal approximation to the binomial, we find \(P(X < 59.5)\).
The z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 59.5\), \(\mu = 75\) and \(\sigma\approx7.984\). So \(z=\frac{59.5 - 75}{7.984}=\frac{-15.5}{7.984}\approx - 1.94\).

Step3: Find the probability from the standard normal table

We look up the value of \(P(Z < - 1.94)\) in the standard normal table. \(P(Z < - 1.94)=0.0262\approx0.0260\) (after rounding).

Answer:

\(0.0260\)