Bicycle sales

A buyer for a department store estimates that the number of mountain bicycles sold during the summer season will range between 40 and 90 with the distribution - number of bicycles sold - 40, 50, 60, 70, 80, 90 with the probability of sale - 0.05, 0.15, 0.4, 0.35, 0.04, 0.01. What is the expected quantity of bicycles sold and with what variance? If 65 bicycles are ordered, what is the probability that some will remain unsold?

Final Answer:

x =  62.1
y =  90.59
p =  0.6

Step-by-step explanation:

$$\begin{gathered} a=40 \\ b=50 \\ c=60 \\ d=70 \\ e=80 \\ f=90 \\ {p}_1=0.05 \\ {p}_2=0.15 \\ {p}_3=0.4 \\ {p}_4=0.35 \\ {p}_5=0.04 \\ {p}_6=0.01 \\ x=a\cdot p_1+b\cdot p_2+c\cdot p_3+d\cdot p_4+e\cdot p_5+f\cdot p_6=40\cdot 0.05+50\cdot 0.15+60\cdot 0.4+70\cdot 0.35+80\cdot 0.04+90\cdot 0.01=62.1 \end{gathered}$$

$y=a^2\cdot p_1+b^2\cdot p_2+c^2\cdot p_3+d^2\cdot p_4+e^2\cdot p_5+f^2\cdot p_6-x^2=40^2\cdot 0.05+50^2\cdot 0.15+60^2\cdot 0.4+70^2\cdot 0.35+80^2\cdot 0.04+90^2\cdot 0.01-62.1^2=90.59$

$$\begin{gathered} p=P(n<65) \\ p=p_1+p_2+p_3=0.05+0.15+0.4=0.6 \end{gathered}$$

Try another example