Respuesta :
Answer:
a) [tex] E(X) = \frac{a+b}{2} = \frac{100+1100}{2}=600[/tex]
b) First we need to calculate the variance given by this formula:
[tex] Var(X) = \frac{(b-a)^2}{12}= \frac{(1100-100)^2}{12} = 83333.33[/tex]
And the deviation would be:
[tex] Sd(X) = \sqrt{83333.33}= 288.675[/tex]
c) [tex] P(600 < X< 889) = P(X<889) -P(X<600)[/tex]
And using the cdf we got:
[tex] P(600 < X< 889)= F(889) -F(600) = \frac{889-100}{1000} -\frac{600-100}{1000}= 0.789- 0.5= 0.289[/tex]
d) [tex] P(311 < X< 889)= F(889) -F(311) = \frac{889-100}{1000} -\frac{311-100}{1000}= 0.789- 0.211= 0.578[/tex]
Step-by-step explanation:
For this case we define the random variable X who represent the customers charge, and the distribution for X on this case is:
[tex] X \sim Unif (a= 100, b=1100)[/tex]
Part a
For this case the average is given by the expected value and we can use the following formula:
[tex] E(X) = \frac{a+b}{2} = \frac{100+1100}{2}=600[/tex]
Part b
First we need to calculate the variance given by this formula:
[tex] Var(X) = \frac{(b-a)^2}{12}= \frac{(1100-100)^2}{12} = 83333.33[/tex]
And the deviation would be:
[tex] Sd(X) = \sqrt{83333.33}= 288.675[/tex]
Part c
For this case we want to find the percent between 600 and 889, so we can use the cumulative distribution function given by:
[tex] F(x) = \frac{X -100}{1100-100}= \frac{x-100}{1000}, 100 \leq X \leq 1100[/tex]
And we can find this probability:
[tex] P(600 < X< 889) = P(X<889) -P(X<600)[/tex]
And using the cdf we got:
[tex] P(600 < X< 889)= F(889) -F(600) = \frac{889-100}{1000} -\frac{600-100}{1000}= 0.789- 0.5= 0.289[/tex]
Part d
[tex] P(311 < X< 889) = P(X<889) -P(X<311)[/tex]
And using the cdf we got:
[tex] P(311 < X< 889)= F(889) -F(311) = \frac{889-100}{1000} -\frac{311-100}{1000}= 0.789- 0.211= 0.578[/tex]
Using the uniform distribution, it is found that:
a) The average charge would be $600.
b) The standard deviation of the monthly amount charged would be $288.68.
c) The percent of monthly charges between $600 and $889 would be 28.9%.
d) The percent of monthly charges between $311 and $889 is 57.8%.
The uniform distribution has two bounds, a and b.
A major credit card company has determined that customers charge between $100 and $1100 per month, hence [tex]a = 100, b = 1100[/tex]
Item a:
The mean of the uniform distribution is:
[tex]M = \frac{a + b}{2}[/tex]
Hence:
[tex]M = \frac{100 + 1100}{2} = 600[/tex]
The average charge would be $600.
Item b:
The standard deviation of the uniform distribution is:
[tex]S = \sqrt{\frac{(b - a)^2}{12}}[/tex]
Hence:
[tex]S = \sqrt{\frac{(1100 - 100)^2}{12}} = 288.68[/tex]
The standard deviation of the monthly amount charged would be $288.68.
Item c:
The probability of finding a value between c and d is:
[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]
Hence:
[tex]P(600 \leq X \leq 889) = \frac{889 - 600}{1100 - 100} = 0.289[/tex]
0.289 x 100% = 28.9%
The percent of monthly charges between $600 and $889 would be 28.9%.
Item d:
[tex]P(311 \leq X \leq 889) = \frac{889 - 311}{1100 - 100} = 0.578[/tex]
0.578 x 100% = 57.8%
The percent of monthly charges between $311 and $889 is 57.8%.
To learn more about the uniform distribution, you can take a look at https://brainly.com/question/13889040
