Math / Data & Statistics

Question(1 point) The daily demand $x$ for a certain product is a random variable with the probability density function $f(x) = \frac{6}{8}x(2-x)$ for $0 \le x \le 2$ .
Determine the expected value of demand: 1
Determine the standard deviation of demand: 0.447
Determine the probability that $x$ is within one standard deviation of the mean: 0.430

Studdy Solution

STEP 1

What is this asking? We're diving into the world of product demand, using probability to figure out some key facts about how much of a product people want! Watch out! Don't forget that we're dealing with a specific range for x, from 0 to 2.
This is crucial for our calculations!

STEP 2

1. Verify the probability density function
2. Calculate the expected value (mean) of demand
3. Determine the standard deviation of demand
4. Find the probability within one standard deviation

STEP 3

Alright, let's kick things off by making sure our probability density function (PDF) is legit.
We've got $f(x) = \frac{6}{8}x(2-x)$ for $0 \leq x \leq 2$ .
To be a valid PDF, it needs to be non-negative and integrate to 1 over its domain.
Let's check it out!

STEP 4

First, is it non-negative?
Well, $\frac{6}{8}$ is positive, x is between 0 and 2, and (2-x) is also positive in this range.
So, check!
It's non-negative.

STEP 5

Now, let's integrate it over the domain:
$\int_0^2 \frac{6}{8}x(2-x) dx = \frac{6}{8}\int_0^2 (2x - x^2) dx$

STEP 6

Time to solve this integral:
$\frac{6}{8}[x^2 - \frac{x^3}{3}]_0^2 = \frac{6}{8}[(4 - \frac{8}{3}) - (0 - 0)] = \frac{6}{8} \cdot \frac{4}{3} = 1$

STEP 7

Look at that!
It integrates to 1.
We've got ourselves a valid PDF.
Let's move on to the good stuff!

STEP 8

Now we're after the expected value, which is the average demand we'd expect to see.
For a continuous random variable, we calculate this using the formula:
$E[X] = \int_a^b x \cdot f(x) dx$
Where a and b are the lower and upper bounds of our domain.

STEP 9

Let's plug in our function and limits:
$E[X] = \int_0^2 x \cdot \frac{6}{8}x(2-x) dx = \frac{6}{8}\int_0^2 (2x^2 - x^3) dx$

STEP 10

Time to integrate:
$E[X] = \frac{6}{8}[\frac{2x^3}{3} - \frac{x^4}{4}]_0^2$

STEP 11

Now let's evaluate this:
$E[X] = \frac{6}{8}[(\frac{16}{3} - 4) - (0 - 0)] = \frac{6}{8} \cdot \frac{4}{3} = 1$

STEP 12

And there we have it!
The expected value of demand is 1.
This matches the answer given in the problem.
On average, we expect the daily demand to be 1 unit.

STEP 13

Next up, we need to find the standard deviation.
This tells us how much the demand typically varies from the mean.
To do this, we first need to calculate the variance, which is the expected value of $(X - E[X])^2$ .
Then we'll take the square root to get the standard deviation.

STEP 14

The formula for variance is:
$Var(X) = E[X^2] - (E[X])^2$
We already know E[X] = 1, so let's calculate E[X^2].

STEP 15

To find E[X^2], we integrate x^2 * f(x):
$E[X^2] = \int_0^2 x^2 \cdot \frac{6}{8}x(2-x) dx = \frac{6}{8}\int_0^2 (2x^3 - x^4) dx$

STEP 16

Let's solve this integral:
$E[X^2] = \frac{6}{8}[\frac{x^4}{2} - \frac{x^5}{5}]_0^2 = \frac{6}{8}[(8 - \frac{32}{5}) - (0 - 0)] = \frac{6}{8} \cdot \frac{8}{5} = \frac{6}{5} = 1.2$

STEP 17

Now we can calculate the variance:
$Var(X) = E[X^2] - (E[X])^2 = 1.2 - 1^2 = 0.2$

STEP 18

The standard deviation is the square root of the variance:
$\sigma = \sqrt{Var(X)} = \sqrt{0.2} \approx 0.447$
This matches the given answer of 0.447!

STEP 19

Now for the grand finale!
We need to find the probability that x is within one standard deviation of the mean.
In other words, we're looking for:
$P(\mu - \sigma \leq X \leq \mu + \sigma)$
Where $\mu$ is the mean (1) and $\sigma$ is the standard deviation (0.447).

STEP 20

Let's calculate our bounds:
$1 - 0.447 = 0.553$ $1 + 0.447 = 1.447$
So we're looking for $P(0.553 \leq X \leq 1.447)$

STEP 21

To find this probability, we need to integrate our PDF between these bounds:
$P(0.553 \leq X \leq 1.447) = \int_{0.553}^{1.447} \frac{6}{8}x(2-x) dx$

STEP 22

Let's solve this integral:
$\frac{6}{8}[\frac{x^2}{2} - \frac{x^3}{3}]_{0.553}^{1.447}$

STEP 23

Evaluating this:
$\frac{6}{8}[(\frac{1.447^2}{2} - \frac{1.447^3}{3}) - (\frac{0.553^2}{2} - \frac{0.553^3}{3})] \approx 0.430$
And look at that!
We've arrived at the answer given in the problem.

STEP 24

The expected value (mean) of demand is 1. The standard deviation of demand is 0.447. The probability that x is within one standard deviation of the mean is 0.430 or about 43%.

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Studdy Solution

STEP 1

What is this asking? We're diving into the world of product demand, using probability to figure out some key facts about how much of a product people want! Watch out! Don't forget that we're dealing with a specific range for x, from 0 to 2.This is crucial for our calculations!

STEP 2

1. Verify the probability density function2. Calculate the expected value (mean) of demand3. Determine the standard deviation of demand4. Find the probability within one standard deviation

STEP 3

STEP 4

First, is it non-negative?Well, 68\frac{6}{8}86​ is positive, x is between 0 and 2, and (2-x) is also positive in this range.So, check!It's non-negative.

STEP 5

Now, let's integrate it over the domain:∫0268x(2−x)dx=68∫02(2x−x2)dx\int_0^2 \frac{6}{8}x(2-x) dx = \frac{6}{8}\int_0^2 (2x - x^2) dx∫02​86​x(2−x)dx=86​∫02​(2x−x2)dx

STEP 6

Time to solve this integral:68[x2−x33]02=68[(4−83)−(0−0)]=68⋅43=1\frac{6}{8}[x^2 - \frac{x^3}{3}]_0^2 = \frac{6}{8}[(4 - \frac{8}{3}) - (0 - 0)] = \frac{6}{8} \cdot \frac{4}{3} = 186​[x2−3x3​]02​=86​[(4−38​)−(0−0)]=86​⋅34​=1

STEP 7

Look at that!It integrates to 1.We've got ourselves a valid PDF.Let's move on to the good stuff!

STEP 8

Now we're after the expected value, which is the average demand we'd expect to see.For a continuous random variable, we calculate this using the formula:E[X]=∫abx⋅f(x)dxE[X] = \int_a^b x \cdot f(x) dxE[X]=∫ab​x⋅f(x)dxWhere a and b are the lower and upper bounds of our domain.

STEP 9

Let's plug in our function and limits:E[X]=∫02x⋅68x(2−x)dx=68∫02(2x2−x3)dxE[X] = \int_0^2 x \cdot \frac{6}{8}x(2-x) dx = \frac{6}{8}\int_0^2 (2x^2 - x^3) dxE[X]=∫02​x⋅86​x(2−x)dx=86​∫02​(2x2−x3)dx

STEP 10

Time to integrate:E[X]=68[2x33−x44]02E[X] = \frac{6}{8}[\frac{2x^3}{3} - \frac{x^4}{4}]_0^2E[X]=86​[32x3​−4x4​]02​

STEP 11

Now let's evaluate this:E[X]=68[(163−4)−(0−0)]=68⋅43=1E[X] = \frac{6}{8}[(\frac{16}{3} - 4) - (0 - 0)] = \frac{6}{8} \cdot \frac{4}{3} = 1E[X]=86​[(316​−4)−(0−0)]=86​⋅34​=1

STEP 12

And there we have it!The expected value of demand is **1**.This matches the answer given in the problem.On average, we expect the daily demand to be 1 unit.

STEP 13

Next up, we need to find the standard deviation.This tells us how much the demand typically varies from the mean.To do this, we first need to calculate the variance, which is the expected value of (X−E[X])2(X - E[X])^2(X−E[X])2.Then we'll take the square root to get the standard deviation.

STEP 14

The formula for variance is:Var(X)=E[X2]−(E[X])2Var(X) = E[X^2] - (E[X])^2Var(X)=E[X2]−(E[X])2We already know E[X] = 1, so let's calculate E[X^2].

STEP 15

To find E[X^2], we integrate x^2 * f(x):E[X2]=∫02x2⋅68x(2−x)dx=68∫02(2x3−x4)dxE[X^2] = \int_0^2 x^2 \cdot \frac{6}{8}x(2-x) dx = \frac{6}{8}\int_0^2 (2x^3 - x^4) dxE[X2]=∫02​x2⋅86​x(2−x)dx=86​∫02​(2x3−x4)dx

STEP 16

STEP 17

Now we can calculate the variance:Var(X)=E[X2]−(E[X])2=1.2−12=0.2Var(X) = E[X^2] - (E[X])^2 = 1.2 - 1^2 = 0.2Var(X)=E[X2]−(E[X])2=1.2−12=0.2

STEP 18

The standard deviation is the square root of the variance:σ=Var(X)=0.2≈0.447\sigma = \sqrt{Var(X)} = \sqrt{0.2} \approx 0.447σ=Var(X)​=0.2​≈0.447This matches the given answer of 0.447!

STEP 19

STEP 20

Let's calculate our bounds:1−0.447=0.5531 - 0.447 = 0.5531−0.447=0.553 1+0.447=1.4471 + 0.447 = 1.4471+0.447=1.447So we're looking for P(0.553≤X≤1.447)P(0.553 \leq X \leq 1.447)P(0.553≤X≤1.447)

STEP 21

To find this probability, we need to integrate our PDF between these bounds:P(0.553≤X≤1.447)=∫0.5531.44768x(2−x)dxP(0.553 \leq X \leq 1.447) = \int_{0.553}^{1.447} \frac{6}{8}x(2-x) dxP(0.553≤X≤1.447)=∫0.5531.447​86​x(2−x)dx

STEP 22

Let's solve this integral:68[x22−x33]0.5531.447\frac{6}{8}[\frac{x^2}{2} - \frac{x^3}{3}]_{0.553}^{1.447}86​[2x2​−3x3​]0.5531.447​

STEP 23

STEP 24

The expected value (mean) of demand is **1**. The standard deviation of demand is **0.447**. The probability that x is within one standard deviation of the mean is **0.430** or about 43%.

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What is this asking? We're diving into the world of product demand, using probability to figure out some key facts about how much of a product people want! Watch out! Don't forget that we're dealing with a specific range for x, from 0 to 2.
This is crucial for our calculations!

1. Verify the probability density function
2. Calculate the expected value (mean) of demand
3. Determine the standard deviation of demand
4. Find the probability within one standard deviation

First, is it non-negative?
Well, $\frac{6}{8}$ is positive, x is between 0 and 2, and (2-x) is also positive in this range.
So, check!
It's non-negative.

Now, let's integrate it over the domain:
$\int_0^2 \frac{6}{8}x(2-x) dx = \frac{6}{8}\int_0^2 (2x - x^2) dx$

Time to solve this integral:
$\frac{6}{8}[x^2 - \frac{x^3}{3}]_0^2 = \frac{6}{8}[(4 - \frac{8}{3}) - (0 - 0)] = \frac{6}{8} \cdot \frac{4}{3} = 1$

Look at that!
It integrates to 1.
We've got ourselves a valid PDF.
Let's move on to the good stuff!

Now we're after the expected value, which is the average demand we'd expect to see.
For a continuous random variable, we calculate this using the formula:
$E[X] = \int_a^b x \cdot f(x) dx$
Where a and b are the lower and upper bounds of our domain.

Let's plug in our function and limits:
$E[X] = \int_0^2 x \cdot \frac{6}{8}x(2-x) dx = \frac{6}{8}\int_0^2 (2x^2 - x^3) dx$

Time to integrate:
$E[X] = \frac{6}{8}[\frac{2x^3}{3} - \frac{x^4}{4}]_0^2$

Now let's evaluate this:
$E[X] = \frac{6}{8}[(\frac{16}{3} - 4) - (0 - 0)] = \frac{6}{8} \cdot \frac{4}{3} = 1$

And there we have it!
The expected value of demand is 1.
This matches the answer given in the problem.
On average, we expect the daily demand to be 1 unit.

Next up, we need to find the standard deviation.
This tells us how much the demand typically varies from the mean.
To do this, we first need to calculate the variance, which is the expected value of $(X - E[X])^2$ .
Then we'll take the square root to get the standard deviation.

The formula for variance is:
$Var(X) = E[X^2] - (E[X])^2$
We already know E[X] = 1, so let's calculate E[X^2].

To find E[X^2], we integrate x^2 * f(x):
$E[X^2] = \int_0^2 x^2 \cdot \frac{6}{8}x(2-x) dx = \frac{6}{8}\int_0^2 (2x^3 - x^4) dx$

Now we can calculate the variance:
$Var(X) = E[X^2] - (E[X])^2 = 1.2 - 1^2 = 0.2$

The standard deviation is the square root of the variance:
$\sigma = \sqrt{Var(X)} = \sqrt{0.2} \approx 0.447$
This matches the given answer of 0.447!

Let's calculate our bounds:
$1 - 0.447 = 0.553$ $1 + 0.447 = 1.447$
So we're looking for $P(0.553 \leq X \leq 1.447)$

To find this probability, we need to integrate our PDF between these bounds:
$P(0.553 \leq X \leq 1.447) = \int_{0.553}^{1.447} \frac{6}{8}x(2-x) dx$

Let's solve this integral:
$\frac{6}{8}[\frac{x^2}{2} - \frac{x^3}{3}]_{0.553}^{1.447}$

The expected value (mean) of demand is 1. The standard deviation of demand is 0.447. The probability that x is within one standard deviation of the mean is 0.430 or about 43%.