Question

A company is considering implementing one of two quality control plans for monitoring the weights of automobile batteries that it manufactures. If the manufacturing process is working properly, the battery weights are approximately normally distributed with a specified mean and standard deviation.

Quality control plan A calls for rejecting a battery as defective if its weight falls more than 2 standard deviations below the specified mean.

Quality control plan B calls for rejecting a battery as defective if its weight falls more than 1.5 interquartile ranges below the lower quartile of the specified population.

Assume the manufacturing process is under control.
a) What proportion of batteries will be rejected by plan A?
b) What is the probability that at least 1 of 2 randomly selected batteries will be rejected by plan A?
c) What proportion of batteries will be rejected by plan B?

Answer 1

@jigglypuff314 @Michele_Laino @nincompoop @robtobey

Answer 2

for questions a) and c), we have to use the tables about the gaussian distribution

Answer 3

have you got those tables?

Answer 4

http://www.pavementinteractive.org/wp-content/uploads/2007/08/Normal_table.gif this?

Answer 5

ok!

Answer 6

now what

Answer 7

Now, in order to answer to question a), you have to look at the row, with z=-2.0 and pick the number at the first column

Answer 8

what do you get?

Answer 9

.0228

Answer 10

ok! Then, the requested probability is:
2.28%

Answer 11

ok! then what do we do

Answer 12

Now, in order to answer to question b), we have to lokk at the row with z= -1.5, and then we have to pick the number at the first column

Answer 13

oopss. question c)

Answer 14

wait so thats the answer for a?

Answer 15

no, since we have a different value for z

Answer 16

wait so we haven't figured out a yet?

Answer 17

I think that the requested proportion, is:
\[\large \frac{{2.28}}{{100}} \cong \frac{2}{{100}}\]

Answer 18

im confused

Answer 19

using the plan A, we reject 2 batteries every 100 bateries produced

Answer 20

oops.. batteries*

Answer 21

oh okay so 2 out of every 100 batteries are rejected for A

Answer 22

yes!

Answer 23

okay :)

Answer 24

now, for questio0n c), we have to lookk at row with z=-1.5, and we have to pick the number at the first column, similarly for question a)

Answer 25

look*

Answer 26

0.0668

Answer 27

ok! So we can say that since we have:
\[\large \frac{{6.68}}{{100}} \cong \frac{7}{{100}}\]
then 7 batteries out for every 100 batteries are rejected for plan B

Answer 28

so for C we say for every 100 batteries for plan B 7 batteries are rejected

Answer 29

yes!

Answer 30

okay now b? :)

Answer 31

please wait I'm trying to figure out that question...

Answer 32

thats fine!

Answer 33

Sincerely speaking, I don't know that answer

Answer 34

Nevertheless, I will continue to try to solve that question

Answer 35

no its fine i think i got it! could you confirm a different question instead?

Answer 36

ok!

Answer 37

this is the question
A GE light bulb is supposed to last for 1200 hours. In fact, light bulbs of this type last only 1185 hours with a standard deviation of 70 hours. What is the probability that a sample of 100 light bulbs will have an average life of at least 1200 hours?

Answer 38

this my work/ answer
70/sqrt100 = 7
P(xbar>1200)=P(xbar>1200)=(Pz>(1200-1185)/7) = 2.142
normalcdf(1200,1E99,1185,7) = -.9839377721
1+ -.9839377721 = 0.160622279

Answer 39

yes! your work is correct, since you have weel apllied the definition of the meaning of the gaussian curve

Answer 40

oops..well*

Answer 41

please wait, the standard deviation is 70, not 7

Answer 42

but don't you do 70/sqt100 = 7?

Answer 43

you are right, yes we have to keep the standard deviation of the mean

Answer 44

ok! your answer is right!
more explanation:
the probability that a light bulb will have an average life of at least of 1220 hours is:
50-48.34= 1.66%

Answer 45

oops..1200 hours

Answer 46

if i submit just what i said is that good or should i add your further explanation

Answer 47

I think that would be better if you express your result as a percentage, since your questions requests a probability. Generally, a probability is expressed in percentage form

Answer 48

so approx 16%

Answer 49

please, wait, I got this:
the probability is:
50-48.38=1.62 approximately

Answer 50

namely 1.62%

Answer 51

where did you get the 50 and 48.32