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Homework 4 Due March 31,2005
1. Page 95 3-100
(a)
P[X = 5] =
(b)
mean = 6.4
P[X = 8] =
(c) mean =9.6
e −3.2 (3.2)5
= .1140
5!
e −6.4 (6.4)8
= .1160
8!
P[x = 0] = e −9.6 = 6.773x10−5
2. Page 99 3-116
(a) Mean # of calls in 30 min = 3
P(X > 3) = 1− P(X ≤ 3) = 1− e−3 − 3e− 3 −
(b)
P[X = 0] = e −3 = .0498
(c)
The mean # of calls in x hours is 6x
e−6x = .01
−6x = ln.01
6x = ln100
ln100
x=
= .7675
6
(d)
e−12 = 6.144x10−6
(e)
e−12 = 6.114x10−12
9e −3 27e− 3
−
= .353
2!
3!
3. Suppose that 10% of all steel shafts produced by a certain process are
nonconforming but can be reworked rather than having to be scrapped.
Consider a random sample of 300 shafts . Use the normal approximation
to the binomial distribution with the continuity correction to find the
probability that the number of nonconforming shafts that can be
reworked is more than 25 but at most 35.
µ = 300(.10) = 30
σ = 300(.10)(.90) = 5.196
25.5 − 30
35.5 − 30
P(25 < x ≤ 35) = P
<Z ≤
 5.196
5.196 
= P(−.87 < Z ≤ 1.06) = .855428 −.19215 = .663278
4. A manufacturing process is designed to produce bolts with a 0.5-in
diameter. Once each day a random sample of 36 bolts is selected and the
diameter is recorded. If the resulting sample mean is less than 0.49 in or
greater than 0.51 inches the process is shut down for readjustment. The
standard deviation of the diameter is .02. What is the probability that the
manufacturing line will be shut down unnecessarily? (Hint: Find the
probability of observing an x in the shutdown range when the true
process mean really is 0.5 in)
.49 − .50
.51− .50 


P(x < .49) + P(x > .51) = P x <
+P x >


.02/6 
.02/ 6 
= P(Z < −3)+ P(Z > 3) = 2(.001350) = .0027