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LenaWriter [7]
2 years ago
5

From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one def

ective transistor, 3% contain two defective transistors, and 2% contain three defective transistors. Calculate the mean and variance for the defective transistors. Mean = Variance = (Please round answers to 4 decimal places.)
Mathematics
1 answer:
jasenka [17]2 years ago
8 0

Answer:

E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500

In order to find the variance we need to find first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300

The variance is calculated with this formula:

Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075

And the standard deviation is just the square root of the variance and we got:

Sd(X) = \sqrt{0.3075}= 0.5545

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X         0           1           2         3

P(X)    0.92     0.03    0.03     0.02

We can calculate the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i)

And replacing we got:

E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500

In order to find the variance we need to find first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300

The variance is calculated with this formula:

Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075

And the standard deviation is just the square root of the variance and we got:

Sd(X) = \sqrt{0.3075}= 0.5545

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Answer/Step-by-step explanation:

Equation to represent the daily rental cost for each type of truck can be written as follows:

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To determine the number of miles for which the truck cost the same amount, set both equations equal to each other and solve for m.

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Check:

Daily rental cost for Trucks-A-Lot = 42 + 0.72m

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Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
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Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

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