Question

decide whether you can use the normal distribution to approximate the binomial distribution. if you can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. if you cannot, explain why and use the binomial distribution to find the indicated probabilities.
a survey of adults found that 9% say their favorite sport is auto racing. you randomly select 600 adults and ask them to name their favorite sport. complete parts (a) through (d).
(b) find the probability that the number of people who say auto racing is their favorite sport is more than 66.
(round to four decimal places as needed.)

Explanation:

Step1: Check if normal approximation is valid

For a binomial distribution \(X\sim B(n,p)\), we can use normal approximation \(X\sim N(np,np(1 - p))\) if \(np\geq5\) and \(n(1 - p)\geq5\).
Here \(n = 600\), \(p=0.09\), \(np=600\times0.09 = 54\geq5\), \(n(1 - p)=600\times(1 - 0.09)=600\times0.91 = 546\geq5\).
So we can use normal approximation \(X\sim N(\mu = np=54,\sigma=\sqrt{np(1 - p)})\).
Calculate \(\sigma=\sqrt{600\times0.09\times(1 - 0.09)}=\sqrt{54\times0.91}=\sqrt{49.14}\approx7.01\).

Step2: Continuity correction and z - score calculation

We want \(P(X>66)\). Using continuity correction for the binomial - to - normal approximation, we find \(P(X>66)\approx P(X>66.5)\) (since for \(P(X > k)\) in binomial, we use \(P(X>k + 0.5)\) in normal approximation).
The z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 66.5\), \(\mu = 54\), \(\sigma\approx7.01\).
\(z=\frac{66.5 - 54}{7.01}=\frac{12.5}{7.01}\approx1.78\).

Step3: Find the probability using the standard normal table

\(P(X>66.5)=1 - P(X\leq66.5)\). From the standard normal table \(P(Z\leq1.78)=0.9625\).
So \(P(X>66.5)=1 - 0.9625=0.0375\).

Answer:

\(0.0375\)