Question

1. nine percent of all men cannot distinguish between the colors red and green. this is the type of color blindness that causes problems with traffic signals. if six men are randomly selected for a study of traffic signal perceptions: (round to 2 decimals.) (a) find the probability that exactly two cannot distinguish between red and green. (b) what are the mean and standard deviation of the random variable x? μ = σ = (c) find the probability that at most two of the men cannot distinguish between red and green. (d) find the probability that more than four of the men cannot distinguish between red and green. (e) in a random sample of six men, two cannot distinguish between the colors red and green. is this result usual or unusual? explain.

Explanation:

Step1: Identify binomial distribution parameters

Let \(n = 6\) (number of trials/sample size), \(p=0.09\) (probability of success, i.e., a man is color - blind), \(q = 1 - p=0.91\)

Step2: Calculate probability for part (a)

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\). For \(k = 2\), \(C(6,2)=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15\). Then \(P(X = 2)=15\times(0.09)^{2}\times(0.91)^{4}\)
\[

$$\begin{align*} P(X = 2)&=15\times0.0081\times0.68574961\\ &=15\times0.005554571841\\ &\approx0.08 \end{align*}$$

\]

Step3: Calculate mean and standard deviation for part (b)

The mean of a binomial distribution is \(\mu=np\), so \(\mu = 6\times0.09 = 0.54\). The standard deviation is \(\sigma=\sqrt{npq}=\sqrt{6\times0.09\times0.91}=\sqrt{0.4914}\approx0.70\)

Step4: Calculate probability for part (c)

\(P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)\)
\(P(X = 0)=C(6,0)\times(0.09)^{0}\times(0.91)^{6}=1\times1\times0.5678393624\approx0.57\)
\(C(6,1)=\frac{6!}{1!(6 - 1)!}=6\), \(P(X = 1)=6\times0.09\times(0.91)^{5}=6\times0.09\times0.623979529\approx0.34\)
\(P(X\leq2)=0.57 + 0.34+0.08=0.99\)

Step5: Calculate probability for part (d)

\(P(X>4)=P(X = 5)+P(X = 6)\)
\(C(6,5)=\frac{6!}{5!(6 - 5)!}=6\), \(P(X = 5)=6\times(0.09)^{5}\times0.91=6\times0.0000059049\times0.91\approx0.000032\)
\(C(6,6)=\frac{6!}{6!(6 - 6)!}=1\), \(P(X = 6)=(0.09)^{6}\approx0.00000053\)
\(P(X>4)\approx0.000032 + 0.00000053\approx0.00\)

Step6: Determine if result is usual or unusual for part (e)

Usual values are within \(\mu\pm2\sigma\). \(\mu - 2\sigma=0.54-2\times0.70=0.54 - 1.4=- 0.86\) (not relevant as \(X\geq0\)), \(\mu + 2\sigma=0.54+2\times0.70=0.54 + 1.4 = 1.94\). Since \(2>1.94\), the result of \(X = 2\) is unusual.

Answer:

(a) \(0.08\)
(b) \(\mu = 0.54\), \(\sigma\approx0.70\)
(c) \(0.99\)
(d) \(0.00\)
(e) Unusual, as \(2\) is outside \(\mu\pm2\sigma\) range (\(\mu + 2\sigma\approx1.94\) and \(2>1.94\))