Question

in a study, 27.4% of individuals in a city regularly use fitness apps. to understand the usage pattern, a fitness company selects a random sample of 37 individuals for a survey. (round your answers to four decimal places.)
a. compute the probability that exactly 16 individuals out of the 37 selected use a fitness app.
b. compute the probability that at most 3 individuals out of the 37 selected use a fitness app.
c. for the sample of 37 individuals, calculate the expected number of users who regularly use fitness apps.
expected value:
d. for the sample of 37 individuals, calculate the variance of the number of users who regularly use fitness apps.
variance:
e. for the sample of 37 individuals, calculate the standard deviation of the number of users who regularly use fitness apps.
standard deviation:

Explanation:

Step1: Identify the binomial distribution parameters

The binomial distribution formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n = 37\) (sample size), \(p=0.274\) (probability of success). For part a, \(k = 16\).
The combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\)
\(C(37,16)=\frac{37!}{16!(37 - 16)!}=\frac{37!}{16!×21!}\)

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\(P(X = 16)=C(37,16)\times(0.274)^{16}\times(1 - 0.274)^{37 - 16}\)

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Using a calculator or statistical software:
\(C(37,16)\approx1.7379\times10^{10}\)
\((0.274)^{16}\approx1.477\times10^{-9}\)
\((0.726)^{21}\approx1.139\times10^{-3}\)
\(P(X = 16)\approx1.7379\times10^{10}\times1.477\times10^{-9}\times1.139\times10^{-3}\approx0.0290\)

Step2: For the expected value (part c)

The formula for the expected value of a binomial distribution is \(E(X)=np\)
Given \(n = 37\) and \(p = 0.274\)
\(E(X)=37\times0.274 = 10.138\)

Step3: For the variance (part d)

The formula for the variance of a binomial distribution is \(V(X)=np(1 - p)\)
Substitute \(n = 37\), \(p = 0.274\)
\(V(X)=37\times0.274\times(1 - 0.274)=37\times0.274\times0.726 = 7.350\)

Step4: For the standard deviation (part e)

The standard deviation \(\sigma=\sqrt{V(X)}\)
Since \(V(X)=7.350\)
\(\sigma=\sqrt{7.350}\approx2.711\)

Step5: For part b (probability that at most 3)

\(P(X\leq3)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\)
\(P(X=k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\)
\(P(X = 0)=\frac{37!}{0!×37!}\times(0.274)^{0}\times(0.726)^{37}=(0.726)^{37}\approx7.97\times10^{-6}\)
\(P(X = 1)=C(37,1)\times(0.274)^{1}\times(0.726)^{36}\)
\(C(37,1)=37\)
\(P(X = 1)=37\times0.274\times(0.726)^{36}\approx37\times0.274\times1.098\times10^{-5}\approx0.00011\)
\(P(X = 2)=C(37,2)\times(0.274)^{2}\times(0.726)^{35}\)
\(C(37,2)=\frac{37!}{2!×35!}=\frac{37\times36}{2\times1}=666\)
\(P(X = 2)=666\times(0.274)^{2}\times(0.726)^{35}\approx666\times0.0751\times1.513\times10^{-5}\approx0.00076\)
\(P(X = 3)=C(37,3)\times(0.274)^{3}\times(0.726)^{34}\)
\(C(37,3)=\frac{37!}{3!×34!}=\frac{37\times36\times35}{3\times2\times1}=7770\)
\(P(X = 3)=7770\times(0.274)^{3}\times(0.726)^{34}\)
\((0.274)^{3}\approx0.0205\)
\((0.726)^{34}\approx2.084\times10^{-5}\)
\(P(X = 3)=7770\times0.0205\times2.084\times10^{-5}\approx0.0033\)
\(P(X\leq3)=7.97\times10^{-6}+ 0.00011+0.00076 + 0.0033\approx0.0042\)

Answer:

a. \(0.0290\)
b. \(0.0042\)
c. \(10.138\)
d. \(7.350\)
e. \(2.711\)