# [Update] Lesson 10: The Binomial Distribution | binomial distribution – Pickpeup

binomial distribution: คุณกำลังดูกระทู้

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. Use the cumulative binomial probability table in the back of your book to find the probability that at most 1 of the 15 sampled has no health insurance.

เนื้อหา

#### 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 \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 1 in the second column on the left, since we want to find \(F(1)=P(X\le 1)\).

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

Table II: continued

p

n
x
0.05
0.10
0.15
0.20
0.25
0.30

11
1.0000
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
1.0000
1.0000
1.0000

14
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

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

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

10
1.0000
1.0000
1.0000
1.0000
0.9999
0.9993

11
1.0000
1.0000
1.0000
1.0000
1.0000
0.9999

12
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

13
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

14
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

15
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

Table II: continued

p

n
x
0.05
0.10
0.15
0.20
0.25
0.30

11
1.0000
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
1.0000
1.0000
1.0000

14
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

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

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

10
1.0000
1.0000
1.0000
1.0000
0.9999
0.9993

11
1.0000
1.0000
1.0000
1.0000
1.0000
0.9999

12
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

13
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

14
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

15
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

We’ve used the cumulative binomial probability table to determine that the probability that at most 1 of the 15 sampled has no health insurance is 0.1671. For kicks, since it wouldn’t take a lot of work in this case, you might want to verify that you’d get the same answer using the binomial p.m.f.

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

#### Solution

As we determined previously, we can calculate \(P(X>7)\) by finding \(P(X\le 7)\) and subtracting it from 1:

0 2 1 3 4 5 6 7 8 9 10 11 12 13 14 15 P(x≤7) P(x≤15)=1 P(x>7)

The good news is that the cumulative binomial probability table makes it easy to determine \(P(X\le 7)\) To find \(P(X\le 7)\) using the binomial table, we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 7 in the second column on the left, since we want to find \(F(7)=P(X\le 7)\).

Now, all we need to do is read the probability value where the \(p=0.20\) column and the (\(n = 15, x = 7\)) row intersect. What do you get?

Table II: continued

p

n
x
0.05
0.10
0.15
0.20
0.25
0.30

11
1.0000
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
1.0000
1.0000
1.0000

14
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

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

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

10
1.0000
1.0000
1.0000
1.0000
0.9999
0.9993

11
1.0000
1.0000
1.0000
1.0000
1.0000
0.9999

12
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

13
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

See also  [Update] UCLA Financial Aid and Scholarships | financial aid and summer classes - Pickpeup

14
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

15
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

Table II: continued

p

n
x
0.05
0.10
0.15
0.20
0.25
0.30

11
1.0000
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
1.0000
1.0000
1.0000

14
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

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

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

10
1.0000
1.0000
1.0000
1.0000
0.9999
0.9993

11
1.0000
1.0000
1.0000
1.0000
1.0000
0.9999

12
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

13
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

14
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

15
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

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

\(P(X>7) = 1 − 0.9958 = 0.0042\)

That is, the probability that more than 7 in a random sample of 15 would have no health insurance is 0.0042.

What is the probability that exactly 3 have no health insurance?

#### Solution

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

0 1 2 3 15 P(x≤2) P(x≤3)

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

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 3 in the second column on the left, since we want to find \(F(3)=P(X\le 3)\). And, find the 2 in the second column on the left, since we want to find \(F(2)=P(X\le 2)\).

Now, all we need to do is (1) read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = 3\)) row intersect, and (2) read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = 2\)) row intersect. What do you get?

Table II: continued

p

n
x
0.05
0.10
0.15
0.20
0.25
0.30

11
1.0000
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
1.0000
1.0000
1.0000

14
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

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

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

10
1.0000
1.0000
1.0000
1.0000
0.9999
0.9993

11
1.0000
1.0000
1.0000
1.0000
1.0000
0.9999

12
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

13
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

14
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

15
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

Table II: continued

p

n
x
0.05
0.10
0.15
0.20
0.25
0.30

11
1.0000
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
1.0000
1.0000
1.0000

14
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

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

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

10
1.0000
1.0000
1.0000
1.0000
0.9999
0.9993

11
1.0000
1.0000
1.0000
1.0000
1.0000
0.9999

12
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

13
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

14
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

15
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

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

\(P(X = 3) = P(X \le 3) – P(X\le 2) = 0.6482-0.3980 = 0.2502\)

That is, there is about a 25% 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 work in this case, you might want to verify that you’d get the same answer using the binomial p.m.f.

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

#### Solution

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

0 1 2 15 P(x≤0) P(x≥1) P(x≤15)=1

To find \(P(X\le 0)\) using the binomial table, we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 0 in the second column on the left, since we want to find \(F(0)=P(X\le 0)\).

Now, all we need to do is read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = 0\)) row intersect. What do you get?

See also  [NEW] Antoni Gaudi | Biography, Sagrada Familia, Works, Buildings, Style, & Facts | gaudi - Pickpeup

Table II: continued

p

n
x
0.05
0.10
0.15
0.20
0.25
0.30

11
1.0000
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
1.0000
1.0000
1.0000

14
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

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

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

10
1.0000
1.0000
1.0000
1.0000
0.9999
0.9993

11
1.0000
1.0000
1.0000
1.0000
1.0000
0.9999

12
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

13
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

14
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

15
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

Table II: continued

p

n
x
0.05
0.10
0.15
0.20
0.25
0.30

11
1.0000
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
1.0000
1.0000
1.0000

14
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

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

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

10
1.0000
1.0000
1.0000
1.0000
0.9999
0.9993

11
1.0000
1.0000
1.0000
1.0000
1.0000
0.9999

12
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

13
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

14
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

15
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

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

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

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 5” means 0, 1, 2, 3, or 4. That is, \(P(X<5)=P(X\le 4)\), and \(P(X\le 4)\) can be readily found using the cumulative binomial table. To find \(P(X\le 4)\), we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 4 in the second column on the left, since we want to find \(F(4)=P(X\le 4)\).

Now, all we need to do is read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = 4\)) row intersect. What do you get?

Table II: continued

p

n
x
0.05
0.10
0.15
0.20
0.25
0.30

11
1.0000
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
1.0000
1.0000
1.0000

14
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

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

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

10
1.0000
1.0000
1.0000
1.0000
0.9999
0.9993

11
1.0000
1.0000
1.0000
1.0000
1.0000
0.9999

12
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

13
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

14
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

15
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

Table II: continued

p

n
x
0.05
0.10
0.15
0.20
0.25
0.30

11
1.0000
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
1.0000
1.0000
1.0000

14
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

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

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

10
1.0000
1.0000
1.0000
1.0000
0.9999
0.9993

11
1.0000
1.0000
1.0000
1.0000
1.0000
0.9999

12
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

13
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

14
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

15
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000

The cumulative binomial probability table tells us that \(P(X\le 4)= 0.8358\). That is, the probability that fewer than 5 people in a random sample of 15 would have no health insurance is 0.8358.

We have now taken a look at an example involving all of the possible scenarios… at most \(x\), more than \(x\), exactly \(x\), at least \(x\), and fewer than \(x\)… of the kinds of binomial probabilities that you might need to find. Oops! Have you noticed that \(p\), the probability of success, in the binomial table in the back of the book only goes up to 0.50. What happens if your \(p\) equals 0.60 or 0.70? All you need to do in that case is turn the problem on its head! For example, suppose you have \(n=10\) and \(p=0.60\), and you are looking for the probability of at most 3 successes. Just change the definition of a success into a failure, and vice versa! That is, finding the probability of at most 3 successes is equivalent to 7 or more failures with the probability of a failure being 0.40. Shall we make this more concrete by looking at a specific example?

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## Binomial distributions | Probabilities of probabilities, part 1

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นอกจากการดูบทความนี้แล้ว คุณยังสามารถดูข้อมูลที่เป็นประโยชน์อื่นๆ อีกมากมายที่เราให้ไว้ที่นี่: ดูเพิ่มเติม

## Binomial Distribution EXPLAINED!

See all my videos at http://www.zstatistics.com/videos/
0:15 Introduction
1:30 Prerequisites/assumptions
2:36 Calculating by hand
8:56 Calculating using Excel
12:05 Expected value and standard deviation
EXAMPLE USED:
Studies show colour blindness affects about 8% of men.
A random sample of 10 men is taken.
Find the probability that:
(a) All 10 men are colour blind
(b) No men are colour blind
(c) Exactly 2 men are colour blind
(d) At least 2 men are colour blind
(e) The expected number of colour blind men in the sample
(f) The standard deviation of the number of
colour blind men in the sample.

## Hàm số – Bài 1 – Toán học 10 – Thầy Lê Thành Đạt (DỄ HIỂU NHẤT)

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## Binomial distribution probability (solve with easy steps)

𝗦𝗵𝗿𝗲𝗻𝗶𝗸 𝗝𝗮𝗶𝗻 𝗦𝘁𝘂𝗱𝘆 𝗦𝗶𝗺𝗽𝗹𝗶𝗳𝗶𝗲𝗱 (𝗔𝗽𝗽 𝗹𝗶𝗻𝗸) :
……………………………………………………
𝗦𝗵𝗿𝗲𝗻𝗶𝗸 𝗝𝗮𝗶𝗻 𝗬𝗼𝘂𝘁𝘂𝗯𝗲 𝗰𝗵𝗮𝗻𝗻𝗲𝗹𝘀
𝗝𝗼𝗯 𝘂𝗽𝗱𝗮𝘁𝗲𝘀 𝗯𝘆 𝗦𝗵𝗿𝗲𝗻𝗶𝗸 𝗝𝗮𝗶𝗻 :
𝗦𝘁𝘂𝗱𝘆𝘀𝗶𝗺𝗽𝗹𝗶𝗳𝗶𝗲𝗱 :
𝗔𝘀𝗸 𝘆𝗼𝘂𝗿 𝗾𝘂𝗲𝗿𝗶𝗲𝘀 :
………………………………………………………………………………..
𝗝𝗼𝗶𝗻 𝗧𝗲𝗹𝗲𝗴𝗿𝗮𝗺 𝗴𝗿𝗼𝘂𝗽 𝘁𝗼 𝗴𝗲𝘁 𝗹𝗮𝘁𝗲𝘀𝘁 𝘂𝗽𝗱𝗮𝘁𝗲𝘀 𝗯𝘆 𝗦𝗵𝗿𝗲𝗻𝗶𝗸 𝗝𝗮𝗶𝗻:
https://t.me/joinchat/Vq6hWaPk5IjHy9Vs
………………………………………………………………………………..
𝗔𝘀𝗸 𝘆𝗼𝘂𝗿 𝗾𝘂𝗲𝘀𝘁𝗶𝗼𝗻 (𝗚𝗼𝗼𝗴𝗹𝗲 𝗳𝗼𝗿𝗺):
https://forms.gle/izmsGkS7j1zN5kN47
……………………………………………………………………………….
𝗦𝗼𝗰𝗶𝗮𝗹 𝘀𝗶𝘁𝗲𝘀 𝗼𝗳 𝗦𝗵𝗿𝗲𝗻𝗶𝗸 𝗝𝗮𝗶𝗻:
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## Finding The Probability of a Binomial Distribution Plus Mean \u0026 Standard Deviation

This Statistics video tutorial explains how to find the probability of a binomial distribution as well as calculating the mean and standard deviation. You need to know how to use the combination formula in order to solve the example problems presented in this video.
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