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For em = 16.50, 2,, = 15152;“ = 2.78.
Thus, the probability is I P (16.20 < .76 < 16.50) = P (1.11 < z < 2.78) = 0.9973  0.8665 = 0.1308. (b) _ _, 15.70—16 __
For :16 —— 15.70, 2 — 0.18 — ——1.67 Thus, the probability is P (z < 15.70) = P(z < —1.67) = 0.475 (4.75%) (C) _ _ 15.20— 6 __
For a: .. 15.20, 2 _ 0181 _ —4.44 Then, the probability is P (I < 15.20) = P (z < —4.44) z 0 (Approximately zero, but not exactly zero) Although it is theoretically possible for a carton to contain less than 15.20 ounces, the probability of this is very close
to zero. Exercise 6.91
(a)
If $3500 2 $1000 + $12w where x is depth in feet then, :1: = 208.33. This means that Company B charges more for depth of more than 208.33 ft..
We have n = 250 and a = 40. For x = 208.33, 2 = 1W = —1.04 Then the probability that Company B charges more than Company A is P (a: > 208.33) = P (z > —1.04) = 1 — 0.1492 = 0.8508.
(b) Using it = 250.
The mean amount charged by Company B is $3500 = $1000 + $1211 = $3500 = $1000 + $12  250 = $4000. Exercise 6.96 We have ,u = 45 and a = 3. Let x be the amount of time Ashley spend commuting to work. The area (under the standard normal curve) to the
left of x has to be 0.95, which gives 2 = 1.65 approximately. Not that we have the relation such that 90—44 Solving the above equation for x, Applying ,u = 45, a = 3, and z = 1.65 1L and H mean "Lower" and "Higher". z = 45+3X1.65
49.50 50 minutes. ll 22 Therefore, Ashley should leave home at about 8 : 10 am in order to arrive at work by 9 am so that she is 95% sure . Exercise 6.98
We have u = 500 but we do not know a. (a) we have P(:r<430) = 0.2
=> P<z=$_”<—430_500)=0.2
U U
:> p(z<é§9ﬂ)=0,2,
(7 Then from the standard normal distribution table, we have 4 _
M : 084.
0
Solving for a,
430 — 500
= ————————— : .3 .
a _084 83 333
(b) 5207'00
For x = 520, z = m = 0.2400.
Thus, P (:1: > 520) = P (z > 02400) = 1 — P (z < 0.2400) = 1 — 0.5948 = 0.4052 (40.52%) Exercise 6.100 Notice that this question is asking the normal distribution approximation of binomial distribution. The original
distributions are binomial. However, some probabilities are computationally hard to calculate under binomial framework.
Thus, we use normal distribution approximation that is computationally easy. However, before we use the approximation
method, we have to check whether or not the conditions for normal approximation are satisﬁed. (See textbook P281). (a) Because the question is asking "What do you think", answers are your guesses. Any answers are ﬁne (Any answers
are acceptable). ('0) Part I: Single—Number Bet (SNB) The probability of winning is __ 1
‘“ 38'
For the normal approximation, we have to check the conditions. (textbook P281) P (SNB Win) 1
np — 25 <38) — 0.6579 < 5.
Thus, one of the conditions is not satisﬁed, and we should NOT use the normal distribution approximation.
Next, calculating the probability of ahead under the framework of binomial distribution. Notice that the gambler can
get positive return (ahead) if he/ she wins more than one time. More precisely,
0 win: Return = —$5 X 25 = —$125 (negative return)
1 wins: Return = $175 X 1 — $5 X 24 = $55 (positive return) 2 wins: Return = $175 X 2 — $5 X 23 = $235 (positive return)
There for the probability of ahead is equivalent to probability of winning more than or equal to one time P (Ahead) 2 P (Winning more than or equal to one time ) Note that the complement of "Winning more than or equal to one time" is "Winning zero time". Then, by applying
the complement rule, we have P (Winning more than or equal to one time ) ll 1 — P (Winning zero time)
= 1— (1 — P(SNB Win))25 25
1— 1—i
( as)
37 25
1_ .—
(38) 0.4866. Part II: Color Bet (CB)
The probability of winning is
18
P CB W' = ~.
( in) 38
For normal approximation, we have to check the conditions. (textbook P281) 18 18 20
1212— 25 (38) — 11.84 > 5 and nq — 25 (1 38) — 25 (28> —— 17.8571 > 5. Thus, we can use the normal approximation. Next, calculating the probability of ahead under the framework of normal distribution. Notice that the gambler can
get positive return (ahead) if he/ she wins more than or equal thirteen time. More precisely, 12 wins: Return = $5 x 12 — $5 x 13 = —$5 (negative return) 13 wins: Return = $5 x 13 — $5 X 12 = $5 (positive return) Now, we use continuity correction (Textbook P282). The positive/negative turning point is 12.5 wins (Not that
this number 12.5 cannot happen in binomial distribution framework since binomial assumes discrete random variables.
However, we are approximating discrete binomial distribution by continuous normal distribution. So, we need to ﬁnd
positive/negative return turning point on the continuous space. That is 12.5 wins). Then, the probability of positive
return (ahead) is P (Ahead) 2 P (x > 12.5) Next, calculating mean and standard deviation. 18
= :2 —— :11. 421
a np 5(38> 8
1820
= ,/—= 25——=2.496r.
‘7 “q 3838 ° Therefore, the probability of positive return is calculated P (:c > 12.5) ll _ (E—M 12.5— 11.8421
P <2 “ a > 2.2465 P (2 > 0.2644)
0.3974 (From normal distribution table)  I Part III: Summarizing
As as result, the (approximated) probabilities of positive return (ahead) are P (Ahead)
P (Ahead) 0.4866 with SingleNumber Bet
0.3974 with Color Bet. Therefore, the gambler has a better chance of coming out positive return (ahead) with the single—number bet. Exercise 6.103 The empirical rule (i) Approximately, 68% of observations fall in the interval p :l: 0 (ii) Approximately, 95% of observations fall in the interval )1. :l: 20 (iii) Approximately, 99.7% of observations fall in the interval M :l: 30 However, above empirical rules are applicable only for Bell—Shape distributions. In this question, we have the binomial distribution with probability p = 0.2, and it is not Bell—Shapedz. So, this question
asks what kinds of problem we will have if we enforce to use normal approximation to Non—Bell—Shaped distribution. There
are some methods to show problems3. Here, I show that normal approximation causes negative random variable that
cannot never happen in a binomial distribution framework. (I believe this is the easiest way to show problems) The mean and standard deviation of this binomial experiment is p = up = 15 (0.2) = 0.3
a = W = V 15 (0.2) (0.8) = 0.5422 Using the empirical rule and obtain the interval where 68% of observations fall. mLawer = [1. + a = 0.3 — (0.5422) 2 —0.2422 (Negative value)
$Upper = M — U = 0.3 + (0.5422) = 0.8422 Therefore, the 68% interval is (—0.2422,0.8422). Here, we ﬁnd the problem. The lower bound of 68% interval is
$1401.,” = —0.2422, the negative value. However, negative random variables never appear in a binomial distribution. We
also can use the empirical rule (ii) and (iii) with the same method, then get negative lower bound values. This means
that if we use the empirical rules, the approximations is problematic. 2Note, if we have a binomial distribution with proability of success close to p = 0.5, it looks like Bell—Shape. Then, normal approximation
is very appropriate method. However, in this question, 1) = 0.2. and it is extremely skewed. Therefore, the normal approximation causes some problems.
3Therefore, this question is an open—end question. ...
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 Spring '07
 Guggenberger

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