Homework Assignment 3 copy
pdf
keyboard_arrow_up
School
New York University *
*We aren’t endorsed by this school
Course
1305
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
Pages
2
Uploaded by JudgeGoldfinch3804
Professor Lakner
COR1-GB 1305.10 V1
5 October 2023
Homework Exercise 3, 4, 9, & 10
3.
A normal random variable
X
has a mean 3.0 and standard deviation 0.2. What is the
probability that
X
falls between 2.75 and 3.1?
P
(2.75 <
X
< 3.1)?
P(
X <
3.1) – P(
X <
2.75)
pnorm(3.1, mean=3.0, sd=0.2) – pnorm(2.75, mean=3.0, sd=0.2)
0.6914625 – 0.1056498 = 0.5858127
4.
Suppose that
X
follows normal distribution with mean 5.5 and standard deviation 0.3. Find a
number
w
such that
X
<
w
with 30% probability.
P
(
X < w
) = .3
qnorm(.3, mean=5.5, sd=0.3)
w
= 5.34268
9.
You are considering a quality inspection scheme to use on the spark plugs which are sent from
your supplier. These spark plugs come in a shipments of 50,000. Denote the unknown proportion
of defective spark plugs in the shipment by
p
. Ideally you would like to reject the shipment if
p
>
.05 and accept it if
p
≤ .05. In practice you can’t follow this plan since you don’t know
p
. Instead
you decide to apply a scheme that consists of the following steps:
A random sample of 20 of the spark plugs will be selected from each shipment. Each of the
selected plugs will be tested to see whether it is defective or not. (The test involves measuring
the plug gap and determining the electrical resistance.) You will note as
X
the (random) number
of defective plugs in the sample. If
X
< 2 then the shipment passes your quality standard. If
X
≥ 2
then the shipment fails the quality test and will be returned to the supplier.
(a)
Find the probability that the shipment is rejected when
p
= .05 (this corresponds to an “error”
since at
p
= .05 we would want to accept the shipment).
P
(
X
≥
2) = 1 –
P
(
X
≤
1)
1 – pbinom(1,20,.05)
1 – 0.7358395 = 0.2641605
(b)
Find the probability that the shipment is accepted when
p
= .1 (this corresponds to an “error”
again since at
p
= .1 we would want to reject the shipment).
P
(
X <
2) =
P
(
X
≤
1)
pbinom(1,20,.1)
0.391747
(c)
Find the probability that the shipment is accepted when
p
= .2.
P
(
X <
2) =
P
(
X
≤
1)
pbinom(1,20,.2)
0.06917529
10.
We would like to modify the quality control test described in question 9 above in the
following way. We want to pass the shipment if
X
<
w
and reject the shipment when
X
≥
w
where
w
is a number to be determined. Determine the smallest possible value for
w
such that the
probability of rejecting the shipment when
p
= .05 is no more than .01, i.e., 1%.
y=0:20
pp=pbinom(y,20,.05)
table=data.frame(y,pp)
table
y
pp
1
0
0.3584859
2
1
0.7358395
3
2
0.9245163
4
3
0.9840985
5
4
0.9974261
6
5
0.9996707
7
6
0.9999661
8
7
0.9999971
9
8
0.9999998
10
9
1.0000000
11
10
1.0000000
12
11
1.0000000
13
12
1.0000000
14
13
1.0000000
15
14
1.0000000
16
15
1.0000000
17
16
1.0000000
18
17
1.0000000
19
18
1.0000000
20
19
1.0000000
21
20
1.0000000
P
(
X
≥
w
)
≤
.01 = P(
X
≤
w
– 1)
≥
.99
P
(
X
≤
4)
≥
.99, so
w
– 1 = 4,
w
= 5
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help