Question

Switzerland is the country with highest consumption of chocolate in the world. In 2017, Swiss consumed almost 9 kg. of chocolate on average. Suppose that a Swiss resident's chocolate consumption can be anywhere between 3 kg, and 15 kg, per year, with equal chance Source: www.statista.com a. What is the probability that a random Swiss resident's annual chocolate consumption is within 4 kg. from the median? The chance that annual consumption is within 4 kg. from the median is: (Round top two decimal places.) b. We are interested in the top 15% of chocolate consumers. What is the interval (x1, x2) in which we can find the annual chocolate consumpt -

Answer 1

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\textup{According to the problem, a resident's chocolate consumption in kgs}=X(\textup{let})\sim\textup U(3,15)$

$f_X(x)=\frac{1}{15-3}=\frac{1}{12}$

$\dpi{150}\textbf{a.}$

Let the median of x be denoted as m.

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\therefore\;\int_3^m \!\!\!\!f_X(x)\,dx=0.5\;\Rightarrow\;\int_3^m\!\!\frac{1}{12}\,dx=0.5\;\Rightarrow\;\frac{1}{12}\int_3^m \!\!\!\!\!\!dx=0.5\;\Rightarrow\;m-3=6\;\Rightarrow\;m=9$

$\textup{Probability that a random resident's chocolate consumption is within 4 kg of the median}$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!=\textup P(9-4\leq X\leq 9+4)=\textup P(5\leq X\leq 13)=\int_5^{13}\!\!\!\!\!\!f_X(x)\,dx=\int_5^{13}\!\!\frac{1}{12}\,dx=\frac{1}{12}\int_5^{13}\!\!\!\!\!\!dx=\frac{1}{12}(8)=\frac 23$

$\dpi{150}\textbf{b.}$

$\textup{The interval (x1,x2) contains the consumption of the top 15}\%\textup{ of the chocolate consumers.}$

$\textup{Thus clearly, x2}=15\textup{ kg}$

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\textup{Thus }:-\;\;\;\;\;\;\;\;\;\;\;\;\int_{\textup{x1}}^{15}\!\!\!\!f_X(x)\,dx=15\%\\\\ \Rightarrow \;\int_{\textup{x1}}^{15}\!\!\frac{1}{12}\,dx=0.15\\\\ \Rightarrow\; \frac{1}{12}\int_{\textup{x1}}^{15}\!\!\!\!\!dx=0.15\\\\ \Rightarrow\;15-\textup x1=1.8\;\Rightarrow\;x1=15-1.8=13.2$

$\textup{Thus x1}=13.2\textup{ kg}$

$\textup{Thus the interval (13.2 kg , 15 kg) contains the consumption of the top }15\%\textup{ chocolate consumers.}$