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a local county has an unemployment rate of 7.1%. a random sample of 19 …

Question

a local county has an unemployment rate of 7.1%. a random sample of 19 employable people are picked at random from the county and are asked if they are employed. round answers to 4 decimal places.
a) find the probability that exactly 2 in the sample are unemployed.
b) find the probability that there are fewer than 3 in the sample are unemployed.
c) find the probability that there are more than 3 in the sample are unemployed.
d) find the probability that there are at most 3 in the sample are unemployed.
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Explanation:

Step1: Identify the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Given \(n = 19\), \(p=0.071\)

Step2: Solve part (a)

For \(k = 2\)
\(C(19,2)=\frac{19!}{2!(19 - 2)!}=\frac{19\times18}{2\times1}=171\)
\(P(X = 2)=171\times(0.071)^{2}\times(1 - 0.071)^{19 - 2}\)
\(P(X = 2)=171\times0.005041\times0.929^{17}\)
\(0.929^{17}\approx0.2997\)
\(P(X = 2)=171\times0.005041\times0.2997\approx0.2578\)

Step3: Solve part (b)

\(P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)\)
For \(k = 0\):
\(C(19,0)=1\)
\(P(X = 0)=1\times(0.071)^{0}\times(0.929)^{19}\approx0.2517\)
For \(k = 1\):
\(C(19,1)=\frac{19!}{1!(19 - 1)!}=19\)
\(P(X = 1)=19\times0.071\times(0.929)^{18}\)
\(0.929^{18}\approx0.2709\)
\(P(X = 1)=19\times0.071\times0.2709\approx0.3677\)
\(P(X\lt3)=0.2517 + 0.3677+0.2578 = 0.8772\)

Step4: Solve part (c)

\(P(X\gt3)=1 - P(X\leq3)\)
\(P(X\leq3)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\)
For \(k = 3\):
\(C(19,3)=\frac{19!}{3!(19 - 3)!}=\frac{19\times18\times17}{3\times2\times1}=969\)
\(P(X = 3)=969\times(0.071)^{3}\times(0.929)^{16}\)
\((0.071)^{3}=0.000357911\)
\(0.929^{16}\approx0.3227\)
\(P(X = 3)=969\times0.000357911\times0.3227\approx0.1119\)
\(P(X\leq3)=0.2517+0.3677 + 0.2578+0.1119=0.9891\)
\(P(X\gt3)=1 - 0.9891=0.0109\)

Step5: Solve part (d)

\(P(X\leq3)=0.9891\) (calculated in part (c))

Answer:

a) \(0.2578\)
b) \(0.8772\)
c) \(0.0109\)
d) \(0.9891\)