Mathematics, 04.03.2020 01:38 tonio638

The probability density function of the weight of packages delivered by a post office is f(x) = 70/69x^2 for 1 < x < 70 pounds.

a) Determine the mean and variance of weight. Round your answers to two decimal places (e. g. 98.76).
Mean = pounds
Variance = pounds2
b) If the shipping cost is $2.50 per pound, what is the average shipping cost of a package? Round your answer to two decimal places (e. g. 98.76).
pounds
c) Determine the probability that the weight of a package exceeds 59 pounds. Round your answer to four decimal places (e. g. 98.7654).

Answers: 2

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Answer from: khynia11

The weights of the package follows a probability density function

The mean is 4.31 and the variance is 51.42, respectively.The average cost of shipping a package is $10.78The probability a package weighs over 59 pounds is 0.0027

The probability density function is given as:

$\mathbf{f(x) = \frac{70}{69x^2},\ 1 < x < 70}$

(a) The mean and the variance

The mean is calculated as:

$\mathbf{E(x) = \int\limits^a_b {x \cdot f(x)} \, dx }$

So, we have:

$\mathbf{E(x) = \int\limits^{70}_1 {x \cdot \frac{70}{69x^2} } \, dx }$

$\mathbf{E(x) = \int\limits^{70}_1 {\frac{70}{69x} } \, dx }$

Rewrite as:

$\mathbf{E(x) = \frac{70}{69}\int\limits^{70}_1 {x^{-1} } \, dx }$

Integrate

$\mathbf{E(x) = \frac{70}{69} {ln(x)}|\limits^{70}_1 } }$

Expand

$\mathbf{E(x) = \frac{70}{69} \cdot {(ln(70) - ln(1)) }}$

$\mathbf{E(x) = 4.31 }}$

The variance is calculated as:

$\mathbf{Var(x) = E(x^2) - (E(x))^2}$

Where:

$\mathbf{E(x^2) = \int\limits^a_b {x^2 \cdot f(x)} \, dx }$

So, we have:

$\mathbf{E(x^2) = \int\limits^{70}_1 {x^2 \cdot \frac{70}{69x^2} } \, dx }$

$\mathbf{E(x^2) = \int\limits^{70}_1 {\frac{70}{69} } \, dx }$

Rewrite as:

$\mathbf{E(x^2) = \frac{70}{69}\int\limits^{70}_1 {1 } \, dx }$

Integrate

$\mathbf{E(x^2) = \frac{70}{69} x|\limits^{70}_1 } }$

Expand

$\mathbf{E(x^2) = \frac{70}{69} \cdot {(70 - 1) }}$

$\mathbf{E(x^2) = 70 }}$

So, we have:

$\mathbf{Var(x) = E(x^2) - (E(x))^2}$

$\mathbf{Var(x) = 70 - 4.31^2}$

$\mathbf{Var(x) = 51.42}$

Hence, the mean is 4.31 and the variance is 51.42, respectively.

(b) The average cost of shipping a package

In (a), we have:

$\mathbf{E(x) = 4.31 }}$ ---- Mean

So, the average cost of shipping a package is:

$\mathbf{Average =4.31 \times 2.50}$

$\mathbf{Average =10.78}$

Hence, the average cost of shipping a package is $10.78

This is represented as: P(x > 59)

So, we have:

$\mathbf{P(x 59) = P(59 < x < 70)}$

So, we have:

$\mathbf{P(x 59) = \int\limits^{70}_{59} { \frac{70}{69x^2}} \, dx }$

Rewrite as:

$\mathbf{P(x 59) = \frac{70}{69}\int\limits^{70}_{59} { x^{-2}} \, dx }$

Integrate

$\mathbf{P(x 59) = \frac{70}{69} \cdot { -\frac 1x}|\limits^{70}_{59}}$

Expand

$\mathbf{P(x 59) = \frac{70}{69} \cdot (-\frac{1}{70} + \frac{1}{59})}$

$\mathbf{P(x 59) = 0.0027}$

Hence, the probability a package weighs over 59 pounds is 0.0027

Read more about probability density functions at:

link

Answer from: aashya16

(a) The mean is 4.31 pounds. The variance is 51.42 pounds.

(b) The average shipping cost of a package is $10.78.

Step-by-step explanation:

The probability density function of the weight of packages is:

$f(x) = \frac{70}{69x^{2}};\ 1 < x < 70$

(a)

The formula for expected value (or mean) of X is:

$E(X)=\int\limits^a_b {x\times f(x)} \, dx$

Compute the expected value of X as follows:

$E(X)=\int\limits^{70}_{1} {x\times \frac{70}{69x^{2}}} \, dx=\frac{70}{69} \int\limits^{70}_{1} {x \times x^{-2}} \, dx\\=\frac{70}{69} \int\limits^{70}_{1} {x^{-1}} \, dx=\frac{70}{69} |\ln x|^{70}_{1}\\=\frac{70}{69}\times\ln 70\\=4.31$

Thus, the mean is 4.31 pounds.

The formula to compute the variance is:

$V(X)=E(X^{2})-[E(X)]^{2}$

Compute the E (X²) as follows:

$E(X^{2})=\int\limits^{70}_{1} {x^{2}\times \frac{70}{69x^{2}}} \, dx=\frac{70}{69} \int\limits^{70}_{1} {x^{2} \times x^{-2}} \, dx\\=\frac{70}{69} \int\limits^{70}_{1} {1} \, dx=\frac{70}{69} | x|^{70}_{1}\\=\frac{70}{69}\times69\\=70$

The variance is:

$V(X)=E(X^{2})-[E(X)]^{2}\\=70-(4.31)^{2}\\=51.4239\\\approx51.42$

Thus, the variance is 51.42 pounds.

(b)

It is provided that the shipping cost for per pound is, C = $2.50.

Compute the average shipping cost of a package as follows:

$Average\ cost=Cost\ per\ pound\times E(X)\\=2.50\times4.31\\=10.775\\\approx10.78$

Thus, the average shipping cost of a package is $10.78.

(c)

Compute the probability that the weight of a package exceeds 59 pounds as follows:

P(59

Thus, the probability that the weight of a package exceeds 59 pounds is 0.0027.

Answer from: Quest

dont understand

step-by-step explanation:

Answer from: Quest

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph

there is a hike in both the graphs so the graphs are increasing on their domain.x-intercept of log x and log 6x are differentthere is a vertical asymptote of both the graphs so there is no y-interceptas both the graphs are increasing so there is no horizontal asymptote

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