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Lyrx [107]
3 years ago
10

Calls to a customer service center last on average 2.8 minutes with a standard deviation of 1.4 minutes. An operator in the call

center is required to answer 75 calls each day. Assume the call times are independent. What is the expected total amount of time in minutes the operator will spend on the calls each day
Mathematics
1 answer:
Natali5045456 [20]3 years ago
6 0

Answer:

The expected total amount of time the operator will spend on the calls each day is of 210 minutes.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

n-values of normal variable:

Suppose we have n values from a normally distributed variable. The mean of the sum of all the instances is M = n\mu and the standard deviation is s = \sigma\sqrt{n}

Calls to a customer service center last on average 2.8 minutes.

This means that \mu = 2.8

75 calls each day.

This means that n = 75

What is the expected total amount of time in minutes the operator will spend on the calls each day

This is M, so:

M = n\mu = 75*2.8 = 210

The expected total amount of time the operator will spend on the calls each day is of 210 minutes.

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Assume that X is normally distributed with a mean of 20 and a standard deviation of 2. Determine the following. (a) P(X 24) (b)
Tems11 [23]

Answer:

a) P( X < 24 ) =  0.9772

b) P ( X > 18 ) =0.8413

c) P ( 14 < X < 26) = 0.9973

d)  P ( 14 < X < 26)  = 0.9973

e) P ( 16 < X < 20)  = 0.4772

f) P ( 20 < X < 26)  =  0.4987

Step-by-step explanation:

Given:

- Mean of the distribution u = 20

- standard deviation sigma = 2

Find:

a. P ( X  < 24 )

b. P ( X  > 18 )

c. P ( 18 < X  < 22 )

d. P ( 14 < X  < 26 )

e. P ( 16 < X  < 20 )

f. P ( 20 < X  < 26 )

Solution:

- We will declare a random variable X that follows a normal distribution

                                   X ~ N ( 20 , 2 )

- After defining our variable X follows a normal distribution. We can compute the probabilities as follows:

a) P ( X < 24 ) ?

- Compute the Z-score value as follows:

                                   Z = (24 - 20) / 2 = 2

- Now use the Z-score tables and look for z = 2:

                                   P( X < 24 ) = P ( Z < 2) = 0.9772

b) P ( X > 18 ) ?

- Compute the Z-score values as follows:

                                   Z = (18 - 20) / 2 = -1

- Now use the Z-score tables and look for Z = -1:

                    P ( X > 18 ) = P ( Z > -1) = 0.8413

c) P ( 18 < X < 22) ?

- Compute the Z-score values as follows:

                                   Z = (18 - 20) / 2 = -1

                                   Z = (22 - 20) / 2 = 1

- Now use the Z-score tables and look for z = -1 and z = 1:

                   P ( 18 < X < 22)  = P ( -1 < Z < 1) = 0.6827

d) P ( 14 < X < 26) ?

- Compute the Z-score values as follows:

                                   Z = (14 - 20) / 2 = -3

                                   Z = (26 - 20) / 2 = 3

- Now use the Z-score tables and look for z = -3 and z = 3:

                   P ( 14 < X < 26)  = P ( -3 < Z < 3) = 0.9973

e) P ( 16 < X < 20) ?

- Compute the Z-score values as follows:

                                   Z = (16 - 20) / 2 = -2

                                   Z = (20 - 20) / 2 = 0

- Now use the Z-score tables and look for z = -2 and z = 0:

                   P ( 16 < X < 20)  = P ( -2 < Z < 0) = 0.4772

f) P ( 20 < X < 26) ?

- Compute the Z-score values as follows:

                                   Z = (26 - 20) / 2 = 3

                                   Z = (20 - 20) / 2 = 0

- Now use the Z-score tables and look for z = 0 and z = 3:

                   P ( 20 < X < 26)  = P ( 0 < Z < 3) = 0.4987

8 0
3 years ago
There are 180 girls in a mixed school. lf the ratio of girls is to boys is 4:3 , the total number of students in the school is w
Angelina_Jolie [31]

Answer:

315

Step-by-step explanation:

No. of girls be g

No. of boys be b

g= 180

g/b = 4/3

b = 3g/4

b= 3(180/4)

b= 3*45

b= 135

.

Find b+g

= 135 + 180

= 315

3 0
3 years ago
D 5/3<br><br><br><br> Couldn’t fit in pic
labwork [276]

Answer:

B

Step-by-step explanation:

You can just move upper y equation and it will be:

6x+3(3x-5)=15

6x+9x-15=15

15x-15=15

15x=30

x=2

6 0
2 years ago
Read 2 more answers
PLS HELP FAST I NEED IT
Gemiola [76]

Answer:

(b) 1:9

(c) 1:8

Step-by-step explanation:

(b) x*y : kx*ky, so 1:k² with k=3 is 1:9

(c) Assuming the rectangles get a z dimension, the volumes would have a ratio of xyz : xkykzk = xyz : xyzk³ = 1 : k³. With k=2 that is 1:8. But the z was never introduced so this is a bit inconclusive.

3 0
3 years ago
A) How many ways can 2 integers from 1,2,...,100 be selected
Anna007 [38]

Answer with explanation:

→Number of Integers from 1 to 100

                                            =100(50 Odd +50 Even)

→50 Even =2,4,6,8,10,12,14,16,...............................100

→50 Odd=1,3,,5,7,9,..................................99.

→Sum of Two even integers is even.

→Sum of two odd Integers is odd.

→Sum of an Odd and even Integer is Odd.

(a)

Number of ways of Selecting 2 integers from 50 Integers ,so that their sum is even,

   =Selecting 2 Even integers from 50 Even Integers , and Selecting 2 Odd integers from 50 Odd integers ,as Order of arrangement is not Important, ,

        =_{2}^{50}\textrm{C}+_{2}^{50}\textrm{C}\\\\=\frac{50!}{(50-2)!(2!)}+\frac{50!}{(50-2)!(2!)}\\\\=\frac{50!}{48!\times 2!}+\frac{50!}{48!\times 2!}\\\\=\frac{50 \times 49}{2}+\frac{50 \times 49}{2}\\\\=1225+1225\\\\=2450

=4900 ways

(b)

Number of ways of Selecting 2 integers from 100 Integers ,so that their sum is Odd,

   =Selecting 1 even integer from 50 Integers, and 1 Odd integer from 50 Odd integers, as Order of arrangement is not Important,

        =_{1}^{50}\textrm{C}\times _{1}^{50}\textrm{C}\\\\=\frac{50!}{(50-1)!(1!)} \times \frac{50!}{(50-1)!(1!)}\\\\=\frac{50!}{49!\times 1!}\times \frac{50!}{49!\times 1!}\\\\=50\times 50\\\\=2500

=2500 ways

7 0
3 years ago
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