Question

According to a survey published by ComPsych Corporation, 54% of all workers read e-mail while they are talking on the phone. Suppose that 20% of those who read e-mail while they are talking on the phone write personal "to-do" lists during meetings. Assuming that these figures are true for all workers, if a worker is randomly selected, determine the following probabilities:
a. The worker reads e-mail while talking on the phone and writes personal "to-do" lists during meetings.
b. The worker does not write personal "to-do" lists given that he reads e-mail while talking on the phone.

Answer 1

OK, let the total no. of workers be P okay?

Answer 2

okay? work with me here.

Answer 3

ok

Answer 4

so The worker reads e-mail while talking on the phone AND writes personal "to-do" lists during meetings
the no. of workers who read e-mail while talking on the phone=0.54(P)

Answer 5

agreed?

Answer 6

yup

Answer 7

now let this no. be Q.
Finally, the no. of workers who read e-mail while talking on the phone AND writes personal "to-do" lists during meetings=0.2(Q), right?

Answer 8

ya

Answer 9

so, this no. is 0.2(q)=0.2*0.54*P
=0.108 P
=10.8% of P

Answer 10

u mean 10.8% of p is the answer for a

Answer 11

yup.

Answer 12

Now for b.

Answer 13

ok

Answer 14

Note that if 20% of those who read e-mail while they are talking on the phone write personal "to-do" lists during meetings, 80% of them DO NOT write these to do lists.
So the required no. is 80% of Q
i.e. 80% of 0.54P
=0.8*0.54*P
=0.432P
=43.2% of P.

Answer 15

done?

Answer 16

ya..

Answer 17

cool.

Answer 18

i m confused

Answer 19

why?

Answer 20

what is that 43.2% of p.. u did already put a value of p i.e. 0.54

Answer 21

and plz do c too

Answer 22

No, it's 80% of Q
Q is nothing but 0.54P.

Answer 23

i got one last question for u... if u don't mind...

Answer 24

sure, but did you understand b?

Answer 25

not really

Answer 26

ok, if the line containing "i.e" is line 1, which line do you find confusing?

Answer 27

instead of writing 0.432p can i simply write 0.432

Answer 28

i guess so, if it confuses you.

Answer 29

will that be right

Answer 30

yes.

Answer 31

and c

Answer 32

post it.

Answer 33

Jessica Salas, president of Salas Products, is reviewing the warranty policy for her company's new model of automobile batteries. Life tests performed on a sample of 100 batteries indicated that an average life of 75 months, a standard deviation of 5 months, and a bell shaped battery life distribution. (a) between what two numbers would 95% of the batteries will fall? (b) between what two numbers would 99.7% of the batteries will fall?

Answer 34

this isn't c :P
Post it as a new question.

Answer 35

c) The worker doesnot write personal "to-do" lists and does read e-mailwhile talking on phone.

Answer 36

Hey dude!!

Answer 37

that is b!

Answer 38

thanks...

Answer 39

one last question and i won't bother u

Answer 40

isn't is .54*.2=.108 in first case...simple p(a)p(b)

Answer 41

ya...

Answer 42

hey Akash, can u help me solve other problem

Answer 43

1-.108...ans to part b

Answer 44

0.432 for b, right?

Answer 45

i need ur help..

Answer 46

Akash, u there?

Answer 47

part b is 1-.108 calculate this , it is not equal to .432

Answer 48

Jessica Salas, president of Salas Products, is reviewing the warranty policy for her company's new model of automobile batteries. Life tests performed on a sample of 100 batteries indicated that an average life of 75 months, a standard deviation of 5 months, and a bell shaped battery life distribution. (a) between what two numbers would 95% of the batteries will fall? (b) between what two numbers would 99.7% of the batteries will fall?

Answer 49

can u plz help me with this question?

Answer 50

post each question as seprate

Answer 51

sorry dude...

Answer 52

here is the question...

Answer 53

Jessica Salas, president of Salas Products, is reviewing the warranty policy for her company's new model of automobile batteries. Life tests performed on a sample of 100 batteries indicated that an average life of 75 months, a standard deviation of 5 months, and a bell shaped battery life distribution. (a) between what two numbers would 95% of the batteries will fall? (b) between what two numbers would 99.7% of the batteries will fall?