Respuesta :
Answer:
The mean of the cable length is [tex]\mu=1205[/tex]
The standard deviation of the cable length is [tex]\sigma=2.886[/tex]
Half of the cables lie in the specifications.
Step-by-step explanation:
Point a:
Suppose X is a continuous random variable with probability density function f(x).
The mean of a continuous random variable, denoted as μ or E(X) is
[tex]\mu= E(X)=\int\limits^\infty_{-\infty} {xf(x)} \, dx[/tex]
The standard deviation of X is
[tex]\sigma=\sqrt{\sigma^2}[/tex]
where [tex]\sigma^2[/tex] is the variance of X
[tex]\sigma^2=\int\limits^\infty_{-\infty} {x^2f(x)} \, dx-\mu^2[/tex]
We know that the probability density function of the length of computer cables is
[tex]f(x)=0.1, \:1200<x<1210[/tex]
Applying the above definition of the mean we get
[tex]E(X)=\int\limits^{1210}_{1200} {0.1x} \, dx =1205\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\\\0.1\cdot \int _{1200}^{1210}xdx\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\0.1\left[\frac{x^{1+1}}{1+1}\right]^{1210}_{1200}\\\\Simplify\\\\0.1\left[0.5x^2\right]^{1210}_{1200}\\\\\mathrm{Compute\:the\:boundaries}:\quad \left[0.5x^2\right]^{1210}_{1200}=12050\\\\0.1\cdot \:12050=1205[/tex]
Applying the above definition of the standard deviation we get
First we need to calculate the variance of X
[tex]\sigma^2=\int\limits^{1210}_{1200} {x^2\cdot 0.1} \, dx-\mu^2[/tex]
[tex]\int _{1200}^{1210}0.1x^2dx\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\\\0.1\cdot \int _{1200}^{1210}x^2dx\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\0.1\left[\frac{x^{2+1}}{2+1}\right]^{1210}_{1200}\\\\0.1\left[\frac{1}{3}\cdot x^3\right]^{1210}_{1200}\\\\\mathrm{Compute\:the\:boundaries}:\quad \left[\frac{1}{3}\cdot x^3\right]^{1210}_{1200}=14520333.33\\\\0.1\cdot \:14520333.33=1452033.33[/tex]
[tex]\sigma^2=1452033.33-1205^2=8.33[/tex]
[tex]\sigma=\sqrt{\sigma^2}=\sqrt{8.33}=2.886[/tex]
Point b:
To find what proportion of cables is within specifications you need to:
[tex]P(1195<X<1205)=\int\limits^{1200}_{1195} {0} \, dx + \int\limits^{1205}_{1200} {0.1} \, dx\\P(1195<X<1205)=0+0.5=0.5[/tex]
You can use the fact that the given distribution is uniform distribution(continuous).
The answers are
- The mean of the given cables' length is 1205 millimetres.
- The standard deviation of the given cables is approx 2.89 millimetres.
- The proportion of the given cables' between 1195 and 1205 is 50%
What is the mean and standard deviation of a uniform distribution?
That distribution, for which all the possible values are equally probable is called uniform distribution.
If the distribution is continuous, let it is defined on interval [a,b]
Then, its mean is [tex]\mu = \dfrac{a+b}{2}[/tex]
Its standard deviation is [tex]\sigma = \dfrac{b-a}{\sqrt{12}}[/tex]
Using the above facts to find the needed information
a) Since we have a = 1200, b = 11210, thus,
Mean = [tex]\mu = \dfrac{a+b}{2} = \dfrac{1200 + 1210}{2} = 1205[/tex]
Standard deviation = [tex]\sigma = \dfrac{b-a}{\sqrt{12}} = \dfrac{1210 - 1200}{\sqrt{12}} \approx 2.89[/tex]
b) To get the proportion for the specified range, we will get the probability of occurrence of length of a cable in range (1195, 1205)
Since the probability is distributed uniformly in range [1200, 1210], thus, each unit difference has 0.1 probability of occurrence.
The range is of 1205-1195 = 10 units, but since the probability for 1195 to 1200 is 0, thus, there are only 5 units valid. Thus 5 times 0.1 = 0.5 probability. Thus, 0.5 or 50/100 or 50% proportion of lengths of cables will lie in that range.
Thus,
The answers are
- The mean of the given cables' length is 1205 millimetres.
- The standard deviation of the given cables is approx 2.89 millimetres.
- The proportion of the given cables' between 1195 and 1205 is 50%
Learn more about uniform distribution here:
https://brainly.com/question/10999921
