Which property is BEST to use when simplifying (0.58+0.32)+ 0.68

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total s

garri49 [273]

Answer

a. The expected total service time for customers = 70 minutes

b. The variance for the total service time = 700 minutes

c. It is not likely that the total service time will exceed 2.5 hours

Step-by-step explanation:

This question is incomplete. I will give the complete version below and proceed with my solution.

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period. (Assume that a sufficient number of servers are available so that no customer must wait for service.) Is it likely that the total service time will exceed 2.5 hours?

Reference

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour.

From the information supplied, we denote that

$X=$ Customers that arrive within the hour

and since $X$ follows a Poisson distribution with mean $\alpha$ = 7

Therefore,

$E(X)= 7$

$& V(X)=7$

Let $Y =$ the total service time for customers arriving during the 1 hour period.

Now, since it takes approximately ten minutes to serve each customer,

$Y=10X$

For a random variable $X$ and a constant $c$ ,

$E(cX)=cE(X)\\V(cX)=c^2V(X)$

Thus,

$E(Y)=E(10X)=10E(X)=10*7=70\\V(Y)=V(10X)=100V(X)=100*7=700$

Therefore the expected total service time for customers = 70 minutes

and the variance for serving time = 700 minutes

Also, the probability of the distribution $Y$ is,

$p_Y(y)=p_x(\frac{y}{10} )\frac{dx}{dy} =\frac{\alpha^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-\alpha } \frac{1}{10}\\ =\frac{7^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-7 } \frac{1}{10}$

So the probability that the total service time exceeds 2.5 hrs or 150 minutes is,

$P(Y>150)=\sum^{\infty}_{k=150} {p_Y} (k) =\sum^{\infty}_{k=150} \frac{7^{\frac{k}{10} }}{(\frac{k}{10})! }.e^{-7} .\frac{1}{10} \\=\frac{7^{\frac{150}{10} }}{(\frac{150}{10})! } .e^{-7}.\frac{1}{10} =0.002$

0.002 is small enough, and the function $\frac{7^{\frac{k}{10} }}{(\frac{k}{10} )!} .e^{-7}.\frac{1}{10}$ gets even smaller when $k$ increases. Hence the probability that the total service time exceeds 2.5 hours is not likely to happen.

3 0

2 years ago

A compound experiment consists of tossing a coin, drawing a letter at random from the alphabet, and then tossing two coins. What

Artemon [7]

We should each take these probabilities first individually.
Probability of tails in a coin toss = 1/2
Probability of drawing letter T from the alphabet = 1/26
Probability of 2 tails in 2 coin toss = 1/4
Sample space = {HH, TT, HT, TH}

Add these together since these events happen after another = 1/2 + 1/26 / + 1/4 = 41/52

The final probability is 41/52

7 0

3 years ago

Read 2 more answers

What percent increase is 21% over 76%

Neporo4naja [7]

Answer:= 261.905% increase

Solution:
Calculate percentage change
from V1 = 21 to V2 = 76
(V2−V1)|V1|×100
(V2−V1)|V1|×100
=(76−21)|21|×100
=(76−21)|21|×100
=5521×100
=5521×100
=2.61905×100
=2.61905×100
=261.905%change
=261.905%change
=261.905%increase

3 0

2 years ago

What is the length of b

oksian1 [2.3K]

Answer:

b = 24

Step-by-step explanation:

Concepts

The Pythagorean Theorem states that the sum of the squares of the two other sides on a right triangle is equal to hypotenuse (longest side) squared. It can be represented by the equation $a^2+b^2=c^2$ . We can also use this formula to solve for the other two legs in the right triangle.