Download 8.8

Section 8.8
This is a section in which we approximate a binomial distribution using a
normal distribution. First we must “judge” whether the approximation is
any good…and that is done by seeing whether BOTH of the following are
true:  - 3 > 0 and  + 3 < 0 (remembering that the mean, , is equal to
n*p and that the standard deviation, , is equal to the square root of n*p*q).
Problem #232
n=10 and p=0.5, so  = 10*0.5 = 5 and  =10*0.5*0.5 = 1.5811
so  - 3 = 5 – 3*1.5811 = 0.2565 which IS greater than zero,
and  + 3 = 5 + 3*1.5811 = 9.7434 which IS less than 10 (n),
so this is a good approximation
Problem #233
n=20 and p=0.4, so  = 20*0.4 = 8 and  =20*0.4*0.6 = 2.1909
so  - 3 = 8 – 3*2.1909= 1.4273 which IS greater than zero,
and  + 3 = 8 + 3*2.1909= 14.5727 which IS less than 20 (n),
so this is a good approximation
Problem #234
n=30 and p=0.2, so  = 30*0.2 = 6 and  =30*0.2*0.8 = 2.1909
so  - 3 = 6 – 3*2.1909= -0.5727 which IS NOT greater than zero,
and  + 3 = 6 + 3*2.1909= 12.5727 which IS less than 30 (n),
so this is NOT a good approximation
Problem #235
n=40 and p=0.1, so  = 40*0.1 = 4 and  =40*0.1*0.9 = 1.8974
so  - 3 = 4 – 3*1.8974= -1.6921 which IS NOT greater than zero,
and  + 3 = 4 + 3*1.8974= 9.6921 which IS less than 40 (n),
so this is NOT a good approximation
Problems #236-239
For these problems, n=20 and p=0.4,
so we will be using  = 20*0.4 = 8 and  =20*0.4*0.6 = 2.1909
Problem #236
The below picture describes the problem…
=8
9
Since this is centered on 9, we must go up a half unit and down a half unit
for the normal distribution, so from 8.5 to 9.5.
Using the same processes that we did in section 8.7, we first get the z-scores:
z8.5= (8.5-8)/2.1909 = 0.23 and z9.5= (9.5-8)/2.1909 = 0.68
…now from the table we get (respectively): 0.0910 and 0.2517;
finally we subtract the two probabilities since they are on the same side of
the mean, giving us: 0.2517 – 0.0910 = 0.1607
Using the table for the exact answer, you get 0.755 -0.596 = 0.169, which is
an OK approximation, but not great
Problem #237
The below picture describes the problem…
=8
12
Since the edge of the shaded area is 12, we must go up a half unit away from
the shaded area for the normal distribution, so 11.5.
Using the same processes that we did in section 8.7, we first get the z-scores:
z11.5= (11.5-8)/2.1909 = 1.60
…now from the table we get: 0.4452;
finally we subtract that from 0.5 since it is a “tail” problem:
0.5 – 0.4452 = 0.0548
Using the table for the exact answer, you get 1 - 0.943 = 0.057, which is a
pretty good approximation.
Problem #238
The below picture describes the problem…
4
=8
9
Since there are two edges to the shaded area, you move a half unit away
from the shaded area, for the normal distribution, so from 3.5 to 9.5.
Using the same processes that we did in section 8.7, we first get the z-scores:
z3.5= (3.5-8)/2.1909 = -2.05 and z9.5= (9.5-8)/2.1909 = 0.68
…now from the table we get (respectively): 0.4798 and 0.2517;
finally we add the two probabilities since they are on the opposite sides of
the mean, giving us: 0.2517 + 0.4798 = 0.7315
Using the table for the exact answer, you get 0.755 -0.016 = 0.739, which is
an OK approximation, but not great
Problem #239
The below picture describes the problem…and the reason for 7 as an end
point is that it is the first number greater than 6…
7
=8
11
Since there are two edges to the shaded area, you move a half unit away
from the shaded area, for the normal distribution, so from 6.5 to 11.5.
Using the same processes that we did in section 8.7, we first get the z-scores:
z6.5= (6.5-8)/2.1909 = -1.14 and z11.5= (11.5-8)/2.1909 = 1.60
…now from the table we get (respectively): 0.3729 and 0.4452;
finally we add the two probabilities since they are on the opposite sides of
the mean, giving us: 0.3729 + 0.4452 = 0.8181
Using the table for the exact answer, you get 0.943 - 0.250 = 0.687, which is
NOT a good approximation
Problem #240
“…fair coin…” means that p equals 0.5; and “…tossed 12 times…” means
that n equals 12…
(a) using the binomial equation: P(3) = C(12,3)*(0.5)3*(0.5)9 = 0.0537
(b) so for the normal approximation, that gives us the following mean and
standard deviation for this problem:
 = 12*0.5 = 6 and  =12*0.5*0.5 = 1.7321
3
=6
Since this is centered on 3, we must go up a half unit and down a half unit
for the normal distribution, so from 2.5 to 3.5.
Using the same processes that we did in section 8.7, we first get the z-scores:
z2.5= (2.5-6)/1.7321 = -2.02 and z3.5= (3.5-6)/1.7321 = -1.44
…now from the table we get (respectively): 0.4783 and 0.4251;
finally we subtract the two probabilities since they are on the same side of
the mean, giving us: 0.4783 – 0.4251 = 0.0532
Problem #241
“…In 84 tosses…” means that n equals 84; and “…points up about 30% of
the time…” means that p equals 0.3… so for the normal approximation, that
gives us the following mean and standard deviation for this problem:
 = 84*0.3 = 25.2 and  =84*0.3*0.7 = 4.2
18
25
Since there are two edges to the shaded area, you move a half unit away
from the shaded area, for the normal distribution, so from 17.5 to 25.5.
Using the same processes that we did in section 8.7, we first get the z-scores:
z17.5= (17.5-25.2)/4.2 = -1.83 and z25.5= (25.5-25.2)/4.2 = 0.07
…now from the table we get (respectively): 0.4664 and 0.0279;
finally we add the two probabilities since they are on the opposite sides of the
mean, giving us: 0.4664 + 0.0279 = 0.4943 (which is only off by 0.0116)
Problem #242
“…6%...is color-blind…” means that p equals 0.06; and “…random sample
of 250…” means that n equals 250… so for the normal approximation, that
gives us the following mean and standard deviation for this problem:
 = 250*0.06 = 15 and  =250*0.06*0.94 = 3.755
10
15
20
Since there are two edges to the shaded area, you move a half unit away
from the shaded area, for the normal distribution, so from 9.5 to 20.5.
Using the same processes that we did in section 8.7, we first get the z-scores:
z9.5= (9.5-15)/3.755 = -1.46 and z20.5= (20.5-15)/3.755 = 1.46
…now from the table we get (respectively): 0.4279 and 0.4279;
finally we add the two probabilities since they are on the opposite sides of the
mean, giving us: 0.4279 + 0.4279 = 0.8558
Problem #243
“…25%...study…” means that p equals 0.25; and “…random sample of
48…” means that n equals 48… so for the normal approximation, that gives
us the following mean and standard deviation for this problem:
 = 48*0.25 = 12 and  =48*0.25*0.75 = 3
12
15
Fifteen is the first number less than 16 (which is one third of 48)…
You move a half unit away from the shaded area, for the normal distribution,
so 15.5.
Using the same processes that we did in section 8.7, we first get the z-scores:
z15.5= (15.5-12)/3= 1.17
…now from the table we get: 0.3790;
finally we add the two probabilities since they are on the opposite sides of the
mean, giving us: 0.5 + 0.3790 = 0.8790
Problem #244
“…60%...favors…” , which means that 40% oppose, so p equals 0.40; and
“…random sample of 96…” means that n equals 96… so for the normal
approximation, that gives us the following mean and standard deviation for
this problem:
 = 96*0.40 = 38.4 and  =96*0.40*0.60 = 4.8
38.4
49
49 is the first number greater than half of 96 (which is majority)…
You move a half unit away from the shaded area, for the normal distribution,
so 48.5.
Using the same processes that we did in section 8.7, we first get the z-scores:
z45.5= (45.5-38.4)/4.8= 2.10
…now from the table we get: 0.4821;
finally since this is “tail” problem, we subtract the probability from 0.5:
0.5 - 0.4821 = 0.0179
Problem #245
“…in 60 trials…” , which means that n equals 60; and “…does not prefer
either shape…” means that p equals 0.5… so for the normal approximation,
that gives us the following mean and standard deviation for this problem:
 = 60*0.50 = 30 and  =60*0.50*0.50 = 3.873
 = 30
40
Since the edge of the shaded area is 40, we must go up a half unit away from
the shaded area for the normal distribution, so 39.5.
Using the same processes that we did in section 8.7, we first get the z-scores:
z39.5= (39.5-30)/3.873 = 2.45
…now from the table we get: 0.4929;
finally we subtract that from 0.5 since it is a “tail” problem:
0.5 – 0.4929 = 0.0071
Problem #246
This is a game problem…so I start with my standard matrix and fill it in…
Event
15+ heads
15+ tails
$
$20
-$10
P($)
$*P($)
The probabilities will have to calculated using a normal approximation…
“…tossed 25 times…” , which means that n equals 25; and since nothing is
stated about the coin’s fairness, it must be assumed to be fair, which means
that p equals 0.5… so for the normal approximation, that gives us the
following mean and standard deviation for this problem:
 = 25*0.50 = 12.5 and  =25*0.50*0.50 = 2.5
 = 12.5
15
Since the edge of the shaded area is 15, we must go up a half unit away from
the shaded area for the normal distribution, so 14.5.
Using the same processes that we did in section 8.7, we first get the z-scores:
z14.5= (14.5-12.5)/2.5 = 0.80
…now from the table we get: 0.2881;
finally we subtract that from 0.5 since it is a “tail” problem:
0.5 – 0.2881 = 0.2119 (which is the probability for 15 or more heads or 15
or more tails, since we assumed a fair coin)…so the game table becomes:
Event
$
P($)
$*P($)
15+ heads
$20
0.2119
$4.24
15+ tails
-$10
0.2119
-$2.12
So the expected value of the game is $4.24 + (-$2.12) = $2.12
Problem #247
“…50 question multiple-choice test…” , which means that n equals 50; and
“…each question having 5 possible answer choices (only one which is
correct)…” means that p equals 1/5 or 0.2… so for the normal
approximation, that gives us the following mean and standard deviation for
this problem:
 = 50*0.20 = 10 and  =50*0.20*0.80 = 2.8284
 = 10
30
Since the edge of the shaded area is 30, we must go up a half unit away from
the shaded area for the normal distribution, so 29.5.
Using the same processes that we did in section 8.7, we first get the z-scores:
z29.5= (29.5-10)/2.8284 = 6.89
…now from the table we get: 0.5;
finally we subtract that from 0.5 since it is a “tail” problem:
0.5 – 0.5 = 0 (not possible)