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you really struggle remembering to bring your lunch to work. each day s…

Question

you really struggle remembering to bring your lunch to work. each day seems to be independent as to whether you remember to bring your lunch or not. the chance that you forget your lunch each day is 21.7%.
consider the next 38 days. let ( x ) be the number of days that you forget your lunch out of the 38 days. calculate the following:
( mu_x ): 8.246
( sigma_x ): 2.541
( p(x = 8) ): 0.149
( p(x < 10) ): 0.697
( p(x > 7) ): 0.636
( p(6 leq x leq 9) ): 0.533
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Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n = 38\), \(p=0.217\), and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Calculate \(P(X = 8)\)

$$ LATEXBLOCK0 $$

Using a calculator, \(P(X = 8)\approx0.163\)

Step3: Calculate \(P(X>7)\)

\(P(X>7)=1 - P(X\leq7)\)

$$ LATEXBLOCK1 $$

Using a binomial - probability calculator (or software like R, Excel), \(P(X\leq7)\approx0.373\), so \(P(X>7)=1 - 0.373 = 0.627\)

Step4: Calculate \(P(6\leq X\leq9)\)

\(P(6\leq X\leq9)=P(X = 6)+P(X = 7)+P(X = 8)+P(X = 9)\)

$$ LATEXBLOCK2 $$

\(P(6\leq X\leq9)=0.119 + 0.147+0.163 + 0.149=0.578\)

Answer:

\(P(X = 8)\approx0.163\), \(P(X>7)\approx0.627\), \(P(6\leq X\leq9)\approx0.578\)