Ask question

Ask question

All categories

2 years ago

5

The probability distribution of the number of loaves of bread sold in a bakery in a week based on past data is given below. Iden

tify the expected number of loaves of bread sold for one week.
Number of Loaves, n: 100 148 135 200
Prob. of n Loaves: 0.25 0.36 0.21 0.18

1 answer:

Tpy6a [65]2 years ago

4 0

Answer:

$E(X) = 100*0.25+ 148*0.36 + 135*0.21 + 200*0.18 = 142.63$

And for this case the expected value for this random variable is given by 142.63 and rounded to the nearest integer we got 143. And that represent the number of loaves of bread sold for one week

Step-by-step explanation:

For this case we have the following distirbution given:

X: 100 148 135 200

P(X): 0.25 0.36 0.21 0.18

The expected value is given by:

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

And replacing we got:

$E(X) = 100*0.25+ 148*0.36 + 135*0.21 + 200*0.18 = 142.63$

And for this case the expected value for this random variable is given by 142.63 and rounded to the nearest integer we got 143. And that represent the number of loaves of bread sold for one week

You might be interested in

Anna bought 8 tetras and 2 rainbow fish for her aquarium. The rainbow fish cost $6 more than the tetras. She paid a total of $

____ [38]

Answer:

C, D, and E

Step-by-step explanation:

When trying to solve for the cost of each item you will find a rainbow fish is $8.50 and a tetra is $2.50. If you use that information to solve the other answer choices C, D, and E would be correct

5 0

2 years ago

Read 2 more answers

Nancy is studying constellations in science class. Two things are strongly supported by all random samples of the number of star

Alona [7]

The first one

2,6,8,3,7,9,4,9,8,7,3,2,8,11,3

7 0

2 years ago

Read 2 more answers

The graph shows the amount of water that remains in a barrel after it begins to leak. The variable x represents the number of da

zhannawk [14.2K]

Answer:

-2

Step-by-step explanation:

i ddid it on edge

5 0

2 years ago

Read 2 more answers

The figure shows the layout of a symmetrical pool in a water park. What is the area of this pool rounded to the tens place? Use

k0ka [10]

Answer:

$2489ft^{2}$

Step-by-step explanation:

The pool are is divided into 4 separated shapes: 2 circular sections and 2 isosceles triangles. Basically, to calculate the whole area, we need to find the area of each section. Due to its symmetry, both triangles are equal, and both circular sections are also the same, so it would be enough to calculate 1 circular section and 1 triangle, then multiply it by 2.

<h3>Area of each triangle:</h3>

From the figure, we know that <em>b = 20ft </em>and <em>h = 25ft. </em>So, the area would be:

$A_{t}=\frac{b.h}{2}=\frac{(20ft)(25ft)}{2}=250ft^{2}$

<h3>Area of each circular section:</h3>

From the figure, we know that $\alpha =2.21 radians$ and the radius is $R=30ft$ . So, the are would be calculated with this formula:

$A_{cs}=\frac{\pi R^{2}\alpha}{360\°}$

Replacing all values:

$A_{cs}=\frac{(3.14)(30ft)^{2}(2.21radians)}{6.28radians}$

Remember that $360\°=6.28radians$

Therefore, $A_{cs}=994.5ft^{2}$

Now, the total are of the figure is:

$A_{total}=2A_{t}+2A{cs}=2(250ft^{2} )+2(994.5ft^{2})\\A_{total}=500ft^{2} + 1989ft^{2}=2489ft^{2}$

Therefore the area of the symmetrical pool is $2489ft^{2}$

3 0

2 years ago

The probability that a call received by a certain switchboard will be a wrong number is 0.02. Use the Poisson distribution to ap

MAXImum [283]

Answer:

0.2008 = 20.08% probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

The probability that a call received by a certain switchboard will be a wrong number is 0.02.

150 calls. So:

$\mu = 150*0.02 = 3$

Use the Poisson distribution to approximate the probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

Either there are less than two calls from wrong numbers, or there are at least two calls from wrong numbers. The sum of the probabilities of these events is 1. So

$P(X < 2) + P(X \geq 2) = 1$

We want to find $P(X \geq 2)$ . So

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0498 + 0.1494 = 0.1992$

Then

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1992 = 0.2008$

0.2008 = 20.08% probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

6 0

2 years ago