Download HW5 on Normal Random Variables Questions:

1
HW5 on Normal Random Variables
.
Name:
Questions:
1. If Z is a standard normal random variable, then P (Z ≤ −2.0) =
.
= 1 − P (Z ≤ 2) = .0228
2. If Z is a standard normal random variable, then P (−1.61 ≤ Z ≤ 1.61) =
.
= Φ(1.61) − Φ(−1.61) = .9463 − .0537 = .8926.
3. In your textbook, the author used the notation zα . z.07 is identical to
th
percentile of standard normal.
93rd percentile.
4. Numerical value of z.17 is
.
Φ(.95) = .83. So the answer is .95.
5. If the population distribution of a variable is approximately normal, then about
%
of the values are within one standard deviation of the mean.
68 %
6. If X is a normally distributed random variable with a mean of 10 and standard deviation of 4, then the probability that X is between 6 and 16 is
.
P (6 ≤ X ≤ 16) = Φ( 16−10
) − Φ( 6−10
) = Φ(1.5) − Φ(−1) = .9332 − .1587 = .7745.
4
4
7. If X is a normally distributed random variable with a mean of 25 and a standard deviation of 8, then the probability that X exceeds 20 is approximately
.
P (X > 20) = 1 − Φ( 20−25
= 1 − Φ(−.625) = 1 − .2643.
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8. If X is a normally distributed random variable with a mean of 80 and a standard
deviation of 12, then 80th percentile of X is
We have Φ( x−80
) = .8, and Φ(.84) = .8 from the table. So
12
that x = 90.08.
.
x−80
12
= .84, and this means
2
9. Suppose X is a normally distributed random variable with unkown mean µ, standard
deviation of 4, and probability that X is less than 5 equal to .7. Then the value of µ
must be
.
70th percentile of X is 5. That means µ + z.3 σ = 5. Since σ = 4, and z.3 = .524, µ
must be 2.904
10. Suppose X is a normally distributed random variable with mean of 25, unknown standard deviation of σ, and probability that X exceeds 27 equal to .25. Then the value
of σ must be
.
75th percentile of X is 27. That means µ + z.25 σ = 27. Since µ = 25 and z.25 = .67, σ
must be 2.985.
11. The distribution of resistance for resistors of a certain type is known to be normal,
with 10% of all resistors having a resistance exceeding 10.634 ohms, and 5% having a
resistance smaller than 9.7565 ohms. What are the mean value and standard deviation
of the resistance distribution?
We have two conditions
P (X < 10.634) = .9
P (X < 9.7565) = .05,
which means
µ + z.1 σ = 10.634
µ + z.95 σ = 9.7565
Since z.1 = 1.28 and z.95 = −1.65, solving the system of two equations, we get σ =
.2995, and µ = 10.2564.
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12. Metal balls for ball bearing are made by machine. Diameter of each ball is known to
have the normal distribution with mean 5 and variance .81 (in mm). Let X = the
diameter of next metal ball to be made.
(a) What is P (X < 6.6)?
) = .9623
Φ( 6.6−5
.9
(b) What is the probability that X is within 2 standard deviation away from 5mm?
apprximately 95% due to 68-95-99.7 rule.
(c) The manager of the plant decides to throw away largest 10% of all the metal balls
made. Give range of metal ball diameter that you should be throwing away.
Largest 10% = µ + z.1 σ. z.1 = 1.28. So the answer is 5 + 1.28 (.9) = 6.152 or
larger.
(d) To make one ball bearing, you need 8 balls having a diameter lager than 4.3mm
and smaller than 5.7mm. The machine can make one ball per minute. What is
the expected time you must wait until you have enough to make one ball bearing?
If you let X =[number of bad balls before you get 8th good one], then X ∼
N B(8, p), where p = P (ball is b/w 4.3 and 5.7). We have
p = P (ball is b/w 4.3 and 5.7) = P (X < 5.7) − P (X < 4.3)
5.7 − 5 4.3 − 5 = Φ
−Φ
.9
.9
5.7 − 5 4.3 − 5 = Φ
−Φ
.9
.9
= .5633
From the formula for Negative Binomial, we have
E(balls made) = E(X + 8) =
8(1 − p)
+ 8 = 14.2
p
So in a long-run average, it takes 14.2 minutes.