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). determine whether a normal distribution can be used to approximate the binomial distribution. choose the correct answer below. a. yes, because both np ≥ 5 and nq ≥ 5. b. no, because np < 5. c. no, because nq < 5. (a) find the probability that the number of people who say auto racing is their favorite sport is at most 63. (round to four decimal places as needed.) sketch the graph of the normal distribution with the indicated probability shaded.

Explanation:

Step1: Check conditions for normal approximation

For a binomial distribution \(X\sim B(n,p)\), we check \(np\) and \(nq\). Here \(n = 600\), \(p=0.09\), \(q = 1 - p=0.91\).
\(np=600\times0.09 = 54\)
\(nq=600\times0.91=546\)
Since \(np\geq5\) and \(nq\geq5\), we can use the normal approximation. The mean of the normal distribution \(\mu=np = 54\), and the standard deviation \(\sigma=\sqrt{npq}=\sqrt{600\times0.09\times0.91}=\sqrt{49.14}\approx7.01\)

Step2: Continuity correction for \(P(X\leq63)\)

For the binomial \(X\) approximated by normal \(Y\sim N(\mu = 54,\sigma\approx7.01)\), when finding \(P(X\leq63)\) for the binomial, using the normal approximation with continuity correction, we find \(P(Y\leq63 + 0.5)=P(Y\leq63.5)\)

Step3: Standardize the value

We use the formula \(z=\frac{x-\mu}{\sigma}\). Substitute \(x = 63.5\), \(\mu = 54\), \(\sigma\approx7.01\)
\(z=\frac{63.5 - 54}{7.01}=\frac{9.5}{7.01}\approx1.36\)

Step4: Find the probability

Using the standard - normal table \(P(Z\leq1.36)\). Looking up the value in the standard - normal table (the cumulative distribution function of the standard normal distribution \(\varPhi(z)\)), we find that \(P(Z\leq1.36)=0.9131\)

Answer:

0.9131