10.3 - Cumulative Binomial Probabilities

Example 10-6 Department

By some estimates, twenty-percentage (20%) of Americans have no health insurance. Randomly sample \(northward=15\) Americans. Permit \(X\) denote the number in the sample with no wellness insurance. What is the probability that exactly three of the 15 sampled have no health insurance?

Solution

Since \(n=15\) is small relative to the population of \(N\) = 300,000,000 Americans, and all of the other criteria pass muster (two possible outcomes, independent trials, ....), the random variable \(X\) tin be assumed to follow a binomial distribution with \(northward=15\) and \(p=0.2\). Using the probability mass office for a binomial random variable, the calculation is then relatively straightforward:

\(P(X=3)=\dbinom{fifteen}{three}(0.20)^iii (0.80)^{12}=0.25\)

That is, there is a 25% chance, in sampling fifteen random Americans, that we would find exactly 3 that had no health insurance.

What is the probability that at well-nigh 1 of those sampled has no health insurance?

Solution

"At most ane" ways either 0 or 1 of those sampled take no health insurance. That is, we need to find:

\(P(10\le 1)=P(X=0)+P(X=1)\)

Using the probability mass part for a binomial random variable with \(northward=15\) and \(p=0.two\), we take:

\(P(X \leq 1)=\dbinom{15}{0}(0.2)^0 (0.8)^{15}+ \dbinom{fifteen}{1}(0.2)^one(0.eight)^{fourteen}=0.0352+0.1319=0.167\)

That is, nosotros have a 16.7% gamble, in sampling 15 random Americans, that we would find at most one that had no health insurance.

What is the probability that more than vii have no health insurance?

Solution

Yikes! "More than seven" in the sample ways 8, nine, 10, eleven, 12, 13, 14, 15. Every bit the following film illustrates, there are two means that we can calculate \(P(X>seven)\):

0 2 1 iii iv v 6 7 8 nine ten 11 12 thirteen 14 fifteen P(x≤seven) P(10≤15)=1 P(10>vii)

We could calculate \(P(X>seven)\) by adding up \(P(Ten=eight), P(X=9)\), up to \(P(X=xv)\). Alternatively, nosotros could calculate \(P(Ten>7)\) by finding \(P(10\le seven)\) and subtracting information technology from 1. But to find \(P(Ten\le 7)\), we'd still accept to add up \(P(X=0), P(X=one)\) upward to \(P(X=7)\). Either way, it becomes readily apparent that answering this question is going to involve more than work than the previous ii questions. Information technology would clearly be helpful if we had an alternative to using the binomial p.chiliad.f. to summate binomial probabilities. The alternative typically used involves cumulative binomial probabilities.

An Bated On Cumulative Probability Distributions Section

Cumulative Probability Distribution

The function:

\(F(x) = P(X\le x)\)

is chosen a cumulative probability distribution. For a discrete random variable \(Ten\), the cumulative probability distribution \(F(ten)\) is determined by:

\(F(10)=\sum\limits_{m=0}^x f(grand)=f(0)+f(1)+\cdots+f(x)\)

Yous'll first desire to annotation that the probability mass function, \(f(x)\), of a detached random variable \(X\) is distinguished from the cumulative probability distribution, \(F(10)\), of a discrete random variable \(X\) by the use of a lowercase \(f\) and an uppercase \(F\). That is, the note f(three) means \(P(X=3)\), while the notation \(F(3)\) means \(P(X\le iii)\).

Now the standard procedure is to report probabilities for a particular distribution as cumulative probabilities, whether in statistical software such as Minitab, a TI-80-something calculator, or in a tabular array similar Table II in the back of your textbook. If y'all take a look at the table, you'll run across that information technology goes on for v pages. Allow's but accept a look at the summit of the kickoff page of the tabular array in guild to go a experience for how the table works:

Solution

In summary, to use the table in the back of your textbook, as well as that found in the back of most probability textbooks, to find cumulative binomial probabilities, exercise the following:

  1. Notice \(n\), the number in the sample, in the first column on the left.
  2. Find the cavalcade containing p, the probability of success.
  3. Find the \(10\) in the second column on the left for which you want to find \(F(ten)=P(X\le x)\).

Let's endeavor it out on our health insurance instance.

Instance x-six Revisited Section

caduceus symbol of medicine

Again, by some estimates, twenty-percent (20%) of Americans have no health insurance. Randomly sample \(n=15\) Americans. Let \(X\) denote the number in the sample with no health insurance. Utilise the cumulative binomial probability tabular array in the back of your volume to find the probability that at most 1 of the fifteen sampled has no health insurance.

Solution

The probability that at most 1 has no health insurance can be written as \(P(X\le 1)\). To find \(P(X\le 1)\) using the binomial table, we:

  1. Find \(northward=15\) in the first column on the left.
  2. Find the cavalcade containing \(p=0.20\).
  3. Notice the 1 in the second column on the left, since we want to find \(F(ane)=P(Ten\le one)\).

At present, all we need to practise is read the probability value where the \(p=0.20\) column and the (\(north=15, x=1\)) row intersect. What do you get?

Table 2: connected
p
n x 0.05 0.x 0.15 0.20 0.25 0.30
eleven i.0000 1.0000 i.0000 1.0000 i.0000 1.0000
12 1.0000 one.0000 i.0000 1.0000 1.0000 1.0000
thirteen ane.0000 i.0000 1.0000 1.0000 1.0000 1.0000
fourteen i.0000 1.0000 1.0000 1.0000 one.0000 1.0000
15 0 0.4633 0.2059 0.0874 0.0352 0.0134 0.0047
1 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353
two 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268
three 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969
iv 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155
v 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216
6 1.0000 0.9997 0.9964 0.9819 0.9434 0.8689
vii one.0000 ane.0000 0.9994 0.9958 0.9827 0.9500
8 1.0000 ane.0000 0.9999 0.9992 0.9958 0.9848
9 one.0000 ane.0000 ane.0000 0.9999 0.9992 0.9963
10 1.0000 ane.0000 one.0000 i.0000 0.9999 0.9993
eleven i.0000 i.0000 i.0000 1.0000 1.0000 0.9999
12 1.0000 1.0000 1.0000 ane.0000 one.0000 1.0000
13 ane.0000 one.0000 i.0000 i.0000 1.0000 1.0000
14 i.0000 1.0000 1.0000 i.0000 i.0000 ane.0000
15 1.0000 1.0000 1.0000 ane.0000 1.0000 i.0000
Table 2: continued
p
n ten 0.05 0.10 0.15 0.20 0.25 0.30
11 one.0000 one.0000 i.0000 1.0000 1.0000 ane.0000
12 1.0000 1.0000 1.0000 one.0000 1.0000 one.0000
13 1.0000 one.0000 ane.0000 i.0000 1.0000 one.0000
14 1.0000 i.0000 one.0000 ane.0000 one.0000 1.0000
fifteen 0 0.4633 0.2059 0.0874 0.0352 0.0134 0.0047
one 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353
2 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268
3 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969
4 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155
v 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216
six i.0000 0.9997 0.9964 0.9819 0.9434 0.8689
7 1.0000 1.0000 0.9994 0.9958 0.9827 0.9500
8 i.0000 i.0000 0.9999 0.9992 0.9958 0.9848
9 1.0000 1.0000 1.0000 0.9999 0.9992 0.9963
10 i.0000 1.0000 i.0000 1.0000 0.9999 0.9993
eleven i.0000 i.0000 1.0000 one.0000 1.0000 0.9999
12 one.0000 1.0000 1.0000 1.0000 1.0000 one.0000
13 1.0000 one.0000 1.0000 1.0000 1.0000 1.0000
fourteen 1.0000 1.0000 1.0000 i.0000 1.0000 i.0000
xv 1.0000 1.0000 1.0000 i.0000 1.0000 1.0000

We've used the cumulative binomial probability tabular array to determine that the probability that at most ane of the fifteen sampled has no wellness insurance is 0.1671. For kicks, since it wouldn't take a lot of work in this case, you might desire to verify that yous'd get the same reply using the binomial p.thou.f.

What is the probability that more than 7 have no wellness insurance?

Solution

Equally we determined previously, we tin can calculate \(P(10>vii)\) past finding \(P(X\le vii)\) and subtracting it from 1:

0 2 1 3 4 5 6 7 eight ix 10 11 12 xiii 14 xv P(x≤7) P(x≤15)=ane P(x>seven)

The proficient news is that the cumulative binomial probability table makes it like shooting fish in a barrel to determine \(P(X\le 7)\) To observe \(P(X\le vii)\) using the binomial table, we:

  1. Detect \(n=xv\) in the showtime cavalcade on the left.
  2. Find the column containing \(p=0.20\).
  3. Discover the 7 in the second column on the left, since we desire to find \(F(7)=P(X\le seven)\).

Now, all we need to do is read the probability value where the \(p=0.20\) column and the (\(n = 15, x = seven\)) row intersect. What do you get?

Table Two: continued
p
n x 0.05 0.10 0.15 0.twenty 0.25 0.thirty
xi 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
12 ane.0000 1.0000 ane.0000 one.0000 1.0000 1.0000
13 1.0000 one.0000 1.0000 1.0000 ane.0000 1.0000
fourteen 1.0000 one.0000 1.0000 1.0000 ane.0000 1.0000
15 0 0.4633 0.2059 0.0874 0.0352 0.0134 0.0047
1 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353
2 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268
3 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969
four 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155
5 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216
half dozen 1.0000 0.9997 0.9964 0.9819 0.9434 0.8689
seven 1.0000 i.0000 0.9994 0.9958 0.9827 0.9500
8 1.0000 1.0000 0.9999 0.9992 0.9958 0.9848
9 one.0000 1.0000 1.0000 0.9999 0.9992 0.9963
10 ane.0000 ane.0000 i.0000 1.0000 0.9999 0.9993
xi one.0000 1.0000 one.0000 1.0000 one.0000 0.9999
12 1.0000 1.0000 1.0000 i.0000 1.0000 1.0000
13 1.0000 1.0000 1.0000 1.0000 one.0000 ane.0000
14 i.0000 1.0000 1.0000 ane.0000 1.0000 one.0000
15 1.0000 ane.0000 ane.0000 1.0000 1.0000 1.0000
Table Two: continued
p
due north x 0.05 0.10 0.15 0.20 0.25 0.30
11 i.0000 i.0000 1.0000 1.0000 1.0000 1.0000
12 1.0000 1.0000 1.0000 i.0000 1.0000 ane.0000
xiii 1.0000 i.0000 1.0000 1.0000 1.0000 i.0000
fourteen 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
15 0 0.4633 0.2059 0.0874 0.0352 0.0134 0.0047
1 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353
ii 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268
iii 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969
4 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155
5 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216
6 1.0000 0.9997 0.9964 0.9819 0.9434 0.8689
7 1.0000 one.0000 0.9994 0.9958 0.9827 0.9500
viii i.0000 ane.0000 0.9999 0.9992 0.9958 0.9848
9 i.0000 ane.0000 1.0000 0.9999 0.9992 0.9963
10 1.0000 1.0000 1.0000 i.0000 0.9999 0.9993
eleven 1.0000 1.0000 1.0000 ane.0000 1.0000 0.9999
12 1.0000 1.0000 1.0000 ane.0000 1.0000 1.0000
thirteen one.0000 1.0000 1.0000 i.0000 1.0000 1.0000
14 1.0000 i.0000 1.0000 1.0000 1.0000 1.0000
15 one.0000 1.0000 1.0000 1.0000 1.0000 1.0000

The cumulative binomial probability table tells us that \(P(Ten\le 7)=0.9958\). Therefore:

\(P(X>seven) = i − 0.9958 = 0.0042\)

That is, the probability that more than vii in a random sample of 15 would have no wellness insurance is 0.0042.

What is the probability that exactly three have no health insurance?

Solution

Nosotros can calculate \(P(X=3)\) by finding \(P(X\le 2)\) and subtracting it from \(P(X\le 3)\), every bit illustrated here:

0 1 2 3 15 P(x≤2) P(ten≤iii)

To find \(P(X\le 2)\) and \(P(Ten\le 3)\) using the binomial table, we:

  1. Find \(north=15\) in the get-go column on the left.
  2. Find the column containing \(p=0.xx\).
  3. Find the three in the second cavalcade on the left, since nosotros want to find \(F(3)=P(X\le 3)\). And, discover the ii in the 2nd cavalcade on the left, since nosotros want to find \(F(ii)=P(10\le 2)\).

Now, all we demand to practise is (i) read the probability value where the \(p = 0.20\) column and the (\(north = 15, ten = 3\)) row intersect, and (ii) read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = two\)) row intersect. What exercise you lot go?

Tabular array II: continued
p
n 10 0.05 0.10 0.15 0.20 0.25 0.xxx
xi 1.0000 ane.0000 1.0000 1.0000 1.0000 1.0000
12 ane.0000 1.0000 i.0000 1.0000 1.0000 1.0000
13 ane.0000 i.0000 1.0000 i.0000 1.0000 ane.0000
xiv 1.0000 1.0000 1.0000 one.0000 1.0000 1.0000
xv 0 0.4633 0.2059 0.0874 0.0352 0.0134 0.0047
1 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353
2 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268
3 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969
4 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155
5 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216
6 1.0000 0.9997 0.9964 0.9819 0.9434 0.8689
7 1.0000 1.0000 0.9994 0.9958 0.9827 0.9500
viii i.0000 1.0000 0.9999 0.9992 0.9958 0.9848
9 one.0000 1.0000 ane.0000 0.9999 0.9992 0.9963
10 1.0000 1.0000 ane.0000 1.0000 0.9999 0.9993
11 1.0000 1.0000 1.0000 1.0000 i.0000 0.9999
12 ane.0000 1.0000 ane.0000 ane.0000 1.0000 1.0000
13 1.0000 one.0000 ane.0000 1.0000 1.0000 1.0000
14 ane.0000 one.0000 1.0000 1.0000 1.0000 1.0000
15 1.0000 1.0000 1.0000 1.0000 one.0000 1.0000
Table Ii: continued
p
n ten 0.05 0.x 0.15 0.xx 0.25 0.30
eleven one.0000 1.0000 1.0000 1.0000 1.0000 1.0000
12 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
13 1.0000 1.0000 1.0000 one.0000 1.0000 1.0000
14 1.0000 1.0000 i.0000 1.0000 1.0000 1.0000
15 0 0.4633 0.2059 0.0874 0.0352 0.0134 0.0047
1 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353
two 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268
3 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969
4 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155
5 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216
6 1.0000 0.9997 0.9964 0.9819 0.9434 0.8689
7 1.0000 i.0000 0.9994 0.9958 0.9827 0.9500
8 1.0000 i.0000 0.9999 0.9992 0.9958 0.9848
9 1.0000 ane.0000 one.0000 0.9999 0.9992 0.9963
10 i.0000 i.0000 1.0000 1.0000 0.9999 0.9993
11 1.0000 1.0000 1.0000 1.0000 ane.0000 0.9999
12 1.0000 ane.0000 1.0000 ane.0000 1.0000 1.0000
13 one.0000 1.0000 i.0000 1.0000 1.0000 1.0000
14 one.0000 1.0000 one.0000 1.0000 1.0000 1.0000
15 i.0000 1.0000 i.0000 1.0000 one.0000 1.0000

The cumulative binomial probability table tells u.s.a. that finding \(P(X\le 3)=0.6482\) and \(P(X\le 2)=0.3980\). Therefore:

\(P(X = 3) = P(10 \le iii) - P(Ten\le 2) = 0.6482-0.3980 = 0.2502\)

That is, there is near a 25% take a chance that exactly 3 people in a random sample of 15 would have no health insurance. Again, for kicks, since it wouldn't take a lot of piece of work in this example, you might desire to verify that you'd get the aforementioned answer using the binomial p.m.f.

What is the probability that at least 1 has no health insurance?

Solution

We can calculate \(P(X\ge ane)\) by finding \(P(X\le 0)\) and subtracting it from 1, equally illustrated here:

0 1 2 15 P(10≤0) P(x≥1) P(ten≤xv)=ane

To notice \(P(X\le 0)\) using the binomial tabular array, nosotros:

  1. Find \(n=15\) in the commencement cavalcade on the left.
  2. Find the cavalcade containing \(p=0.20\).
  3. Find the 0 in the second cavalcade on the left, since nosotros desire to find \(F(0)=P(Ten\le 0)\).

Now, all we need to do is read the probability value where the \(p = 0.twenty\) column and the (\(north = 15, x = 0\)) row intersect. What do you get?

Tabular array II: connected
p
north 10 0.05 0.10 0.xv 0.20 0.25 0.30
xi i.0000 1.0000 1.0000 1.0000 one.0000 i.0000
12 1.0000 1.0000 1.0000 ane.0000 1.0000 i.0000
xiii one.0000 one.0000 1.0000 1.0000 ane.0000 one.0000
fourteen 1.0000 1.0000 1.0000 1.0000 ane.0000 1.0000
15 0 0.4633 0.2059 0.0874 0.0352 0.0134 0.0047
one 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353
2 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268
3 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969
4 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155
v 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216
6 ane.0000 0.9997 0.9964 0.9819 0.9434 0.8689
7 i.0000 ane.0000 0.9994 0.9958 0.9827 0.9500
8 i.0000 1.0000 0.9999 0.9992 0.9958 0.9848
ix one.0000 1.0000 one.0000 0.9999 0.9992 0.9963
10 i.0000 ane.0000 1.0000 one.0000 0.9999 0.9993
11 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999
12 i.0000 1.0000 1.0000 1.0000 ane.0000 one.0000
13 1.0000 one.0000 i.0000 i.0000 ane.0000 1.0000
14 1.0000 1.0000 1.0000 ane.0000 1.0000 1.0000
15 i.0000 1.0000 1.0000 ane.0000 1.0000 1.0000
Table 2: connected
p
northward 10 0.05 0.10 0.15 0.20 0.25 0.xxx
xi one.0000 1.0000 ane.0000 1.0000 ane.0000 1.0000
12 i.0000 one.0000 1.0000 1.0000 1.0000 i.0000
13 1.0000 ane.0000 1.0000 i.0000 one.0000 1.0000
fourteen 1.0000 1.0000 1.0000 1.0000 1.0000 i.0000
xv 0 0.4633 0.2059 0.0874 0.0352 0.0134 0.0047
1 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353
2 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268
iii 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969
iv 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155
5 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216
6 ane.0000 0.9997 0.9964 0.9819 0.9434 0.8689
vii 1.0000 1.0000 0.9994 0.9958 0.9827 0.9500
8 1.0000 1.0000 0.9999 0.9992 0.9958 0.9848
9 1.0000 1.0000 1.0000 0.9999 0.9992 0.9963
ten one.0000 1.0000 1.0000 one.0000 0.9999 0.9993
11 one.0000 i.0000 ane.0000 i.0000 ane.0000 0.9999
12 1.0000 1.0000 i.0000 1.0000 1.0000 1.0000
thirteen one.0000 1.0000 1.0000 1.0000 1.0000 1.0000
14 i.0000 ane.0000 1.0000 1.0000 1.0000 1.0000
15 ane.0000 i.0000 i.0000 1.0000 1.0000 1.0000

The cumulative binomial probability table tells us that \(P(10\le 0)=0.0352\). Therefore:

\(P(X\le 1) = one-0.0352 = 0.9648\)

That is, the probability that at least one person in a random sample of 15 would have no health insurance is 0.9648.

What is the probability that fewer than 5 have no health insurance?

Solution

"Fewer than five" means 0, 1, 2, three, or 4. That is, \(P(X<v)=P(X\le 4)\), and \(P(X\le 4)\) tin can be readily establish using the cumulative binomial tabular array. To find \(P(X\le 4)\), we:

  1. Find \(north=15\) in the starting time cavalcade on the left.
  2. Find the column containing \(p=0.20\).
  3. Observe the iv in the second column on the left, since we want to find \(F(4)=P(Ten\le 4)\).

Now, all nosotros need to do is read the probability value where the \(p = 0.twenty\) column and the (\(north = 15, x = iv\)) row intersect. What do you get?

Table Two: continued
p
n x 0.05 0.10 0.xv 0.20 0.25 0.thirty
11 ane.0000 1.0000 1.0000 ane.0000 1.0000 1.0000
12 1.0000 one.0000 1.0000 1.0000 one.0000 1.0000
13 1.0000 ane.0000 1.0000 i.0000 1.0000 1.0000
14 ane.0000 1.0000 ane.0000 1.0000 i.0000 1.0000
xv 0 0.4633 0.2059 0.0874 0.0352 0.0134 0.0047
1 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353
2 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268
3 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969
iv 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155
5 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216
half dozen 1.0000 0.9997 0.9964 0.9819 0.9434 0.8689
7 1.0000 i.0000 0.9994 0.9958 0.9827 0.9500
8 1.0000 1.0000 0.9999 0.9992 0.9958 0.9848
ix 1.0000 1.0000 1.0000 0.9999 0.9992 0.9963
10 1.0000 1.0000 i.0000 one.0000 0.9999 0.9993
11 one.0000 ane.0000 1.0000 one.0000 i.0000 0.9999
12 1.0000 1.0000 1.0000 1.0000 one.0000 ane.0000
13 1.0000 1.0000 i.0000 1.0000 one.0000 ane.0000
xiv one.0000 1.0000 one.0000 1.0000 1.0000 ane.0000
15 1.0000 1.0000 one.0000 one.0000 one.0000 ane.0000
Tabular array Two: connected
p
n ten 0.05 0.10 0.fifteen 0.xx 0.25 0.30
11 1.0000 one.0000 one.0000 1.0000 1.0000 1.0000
12 1.0000 i.0000 i.0000 1.0000 1.0000 1.0000
thirteen ane.0000 i.0000 1.0000 1.0000 i.0000 ane.0000
14 one.0000 ane.0000 1.0000 1.0000 1.0000 1.0000
15 0 0.4633 0.2059 0.0874 0.0352 0.0134 0.0047
1 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353
2 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268
3 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969
four 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155
five 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216
6 1.0000 0.9997 0.9964 0.9819 0.9434 0.8689
7 i.0000 i.0000 0.9994 0.9958 0.9827 0.9500
eight 1.0000 1.0000 0.9999 0.9992 0.9958 0.9848
ix 1.0000 1.0000 1.0000 0.9999 0.9992 0.9963
10 1.0000 1.0000 1.0000 i.0000 0.9999 0.9993
xi 1.0000 1.0000 ane.0000 1.0000 1.0000 0.9999
12 1.0000 1.0000 i.0000 1.0000 1.0000 1.0000
13 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
xiv 1.0000 1.0000 ane.0000 one.0000 1.0000 1.0000
xv 1.0000 1.0000 ane.0000 1.0000 one.0000 1.0000

The cumulative binomial probability tabular array tells us that \(P(10\le 4)= 0.8358\). That is, the probability that fewer than v people in a random sample of 15 would have no wellness insurance is 0.8358.

Nosotros have now taken a look at an example involving all of the possible scenarios... at well-nigh \(x\), more than \(x\), exactly \(x\), at least \(x\), and fewer than \(x\)... of the kinds of binomial probabilities that you lot might need to notice. Oops! Take you lot noticed that \(p\), the probability of success, in the binomial table in the back of the book merely goes up to 0.50. What happens if your \(p\) equals 0.60 or 0.70? All yous demand to do in that example is turn the problem on its caput! For example, suppose you have \(n=x\) and \(p=0.sixty\), and you are looking for the probability of at nearly iii successes. Only alter the definition of a success into a failure, and vice versa! That is, finding the probability of at well-nigh 3 successes is equivalent to 7 or more than failures with the probability of a failure being 0.40. Shall nosotros make this more than concrete past looking at a specific example?

Example 10-7 Section

electric meter

Many utility companies promote energy conservation by offering discount rates to consumers who keep their free energy usage below certain established subsidy standards. A recent EPA report notes that 70% of the island residents of Puerto Rico accept reduced their electricity usage sufficiently to qualify for discounted rates. If ten residential subscribers are randomly selected from San Juan, Puerto Rico, what is the probability that at least iv qualify for the favorable rates?

Solution

If we let \(10\) announce the number of subscribers who qualify for favorable rates, then Ten is a binomial random variable with \(n=ten\) and \(p=0.seventy\). And, if we allow \(Y\) denote the number of subscribers who don't qualify for favorable rates, then \(Y\), which equals \(10-X\), is a binomial random variable with \(n=10\) and \(q=1-p=0.30\). We are interested in finding \(P(X\ge 4)\). Nosotros can't use the cumulative binomial tables, because they just go up to \(p=0.fifty\). The good news is that we can rewrite \(P(X\ge 4)\)as a probability argument in terms of \(Y\):

\(P(Ten\ge 4) = P(-Ten\le -4) = P(10 -Ten\le ten - 4) = P(Y\le 6)\)

Now information technology's just a affair of looking upwards the probability in the right place on our cumulative binomial table. To find \(P(Y\le 6)\), we:

  1. Discover \(n=x\) in the first column on the left.
  2. Find the column containing \(p=0.xxx\).
  3. Notice the 6 in the second column on the left, since we want to find \(F(vi)=P(Y\le 6)\).

At present, all we need to do is read the probability value where the \(p = 0.30\) column and the (\(n = 10, y = half-dozen\)) row intersect. What do you get?

p
n x 0.05 0.10 0.fifteen 0.20 0.25 0.30 0.35
six i.0000 1.0000 1.0000 i.0000 0.9999 0.9998 0.9994
vii one.0000 i.0000 i.0000 ane.0000 1.0000 i.0000 1.0000
8 0 0.6634 0.4305 0.2725 0.1678 0.1001 0.0576 0.0319
i 0.9428 0.8131 0.6572 0.5033 0.3671 0.2553 0.1691
two 0.9942 0.9619 0.8948 0.7969 0.6785 0.5518 0.4278
iii 0.9996 0.99950 0.9786 0.9437 0.8862 0.8059 0.7064
four i.0000 0.9996 0.9971 0.9896 0.9727 0.9420 0.8939
five 1.0000 1.0000 0.9998 0.9988 0.9958 0.9887 0.9747
6 1.0000 1.0000 one.0000 0.9999 0.9996 0.9987 0.9964
7 1.0000 1.0000 i.0000 1.0000 1.0000 0.9999 0.9998
viii ane.0000 1.0000 ane.0000 1.0000 1.0000 1.0000 1.0000
9 0 0.6302 0.3874 0.2316 0.1342 0.0751 0.0404 0.0207
1 0.9288 0.7748 0.5995 0.4362 0.3003 0.1960 0.1211
2 0.9916 0.9470 0.8591 0.7382 0.6007 0.4625 0.3373
3 0.9994 0.9917 0.9661 0.9144 0.8343 0.7297 0.6089
4 ane.0000 0.9991 0.9944 0.9804 0.9511 0.9012 0.8283
5 1.0000 0.9999 0.9994 0.9969 0.9900 0.9747 0.9464
half-dozen 1.0000 one.0000 1.0000 0.9997 0.9987 0.9957 0.9888
7 1.0000 one.0000 1.0000 one.0000 0.9999 0.9996 0.9986
8 one.0000 i.0000 one.0000 1.0000 1.0000 ane.0000 0.9999
ix 1.0000 1.0000 1.0000 1.0000 ane.0000 ane.0000 ane.0000
x 0 0.5987 0.3487 0.1969 0.1074 0.0563 0.0282 0.0135
i 0.9139 0.7361 0.5443 0.3758 0.2440 0.1493 0.0860
ii 0.9885 0.9298 0.8202 0.6778 0.5256 0.3828 0.2616
three 0.990 0.9872 0.9500 0.8591 0.7759 0.6496 0.5138
iv 0.9999 0.9984 0.9901 0.9672 0.9219 0.8497 0.7515
5 1.0000 0.9999 0.9986 0.9936 0.9803 0.9527 0.9051
6 ane.0000 1.0000 0.9999 0.9991 0.9965 0.9894 0.9740
seven 1.0000 1.0000 1.0000 0.9999 0.9996 0.9984 0.9952
eight one.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9995
9 1.0000 ane.0000 1.0000 1.0000 one.0000 1.0000 1.0000
ten 1.0000 one.0000 1.0000 1.0000 1.0000 i.0000 ane.0000
p
north 10 0.05 0.10 0.fifteen 0.xx 0.25 0.30 0.35
6 1.0000 1.0000 ane.0000 1.0000 0.9999 0.9998 0.9994
7 1.0000 ane.0000 1.0000 ane.0000 1.0000 1.0000 1.0000
viii 0 0.6634 0.4305 0.2725 0.1678 0.1001 0.0576 0.0319
1 0.9428 0.8131 0.6572 0.5033 0.3671 0.2553 0.1691
ii 0.9942 0.9619 0.8948 0.7969 0.6785 0.5518 0.4278
3 0.9996 0.99950 0.9786 0.9437 0.8862 0.8059 0.7064
four ane.0000 0.9996 0.9971 0.9896 0.9727 0.9420 0.8939
5 ane.0000 1.0000 0.9998 0.9988 0.9958 0.9887 0.9747
half-dozen 1.0000 one.0000 1.0000 0.9999 0.9996 0.9987 0.9964
7 i.0000 i.0000 1.0000 1.0000 ane.0000 0.9999 0.9998
eight i.0000 ane.0000 1.0000 1.0000 1.0000 1.0000 ane.0000
ix 0 0.6302 0.3874 0.2316 0.1342 0.0751 0.0404 0.0207
1 0.9288 0.7748 0.5995 0.4362 0.3003 0.1960 0.1211
2 0.9916 0.9470 0.8591 0.7382 0.6007 0.4625 0.3373
3 0.9994 0.9917 0.9661 0.9144 0.8343 0.7297 0.6089
4 1.0000 0.9991 0.9944 0.9804 0.9511 0.9012 0.8283
5 i.0000 0.9999 0.9994 0.9969 0.9900 0.9747 0.9464
6 1.0000 one.0000 1.0000 0.9997 0.9987 0.9957 0.9888
vii 1.0000 1.0000 1.0000 one.0000 0.9999 0.9996 0.9986
8 1.0000 ane.0000 1.0000 1.0000 1.0000 ane.0000 0.9999
9 ane.0000 ane.0000 one.0000 1.0000 i.0000 1.0000 1.0000
x 0 0.5987 0.3487 0.1969 0.1074 0.0563 0.0282 0.0135
1 0.9139 0.7361 0.5443 0.3758 0.2440 0.1493 0.0860
2 0.9885 0.9298 0.8202 0.6778 0.5256 0.3828 0.2616
iii 0.990 0.9872 0.9500 0.8591 0.7759 0.6496 0.5138
4 0.9999 0.9984 0.9901 0.9672 0.9219 0.8497 0.7515
5 1.0000 0.9999 0.9986 0.9936 0.9803 0.9527 0.9051
six one.0000 1.0000 0.9999 0.9991 0.9965 0.9894 0.9740
7 1.0000 1.0000 1.0000 0.9999 0.9996 0.9984 0.9952
8 i.0000 1.0000 one.0000 1.0000 1.0000 0.9999 0.9995
9 1.0000 1.0000 1.0000 1.0000 ane.0000 i.0000 i.0000
10 one.0000 1.0000 1.0000 1.0000 one.0000 1.0000 1.0000

The cumulative binomial probability table tells the states that \(P(Y\le 6)=P(Ten\ge 4)=0.9894\). That is, the probability that at to the lowest degree four people in a random sample of ten would qualify for favorable rates is 0.9894.

If you lot are in demand of calculating binomial probabilities for more specific probabilities of success (\(p\)), such every bit 0.37 or 0.61, you can apply statistical software, such equally Minitab, to determine the cumulative binomial probabilities. You can then still employ the methods illustrated hither on this page to notice the specific probabilities (more than \(ten\), fewer than \(10\), ...) that you need.