Suppose that you have a fair coin and you label the head side as 1 and the tail side as 0. a. Now, you are asked to flip the coin 2 times and write down the numerical value that results from each toss. Without actually flipping the coin, write down the sampling distribution of the sample means. b. Repeat part (a) with the coin flipped 4 times. c. Repeat part (a) with the coin flipped 10 times.

In the explanation

Step-by-step explanation:

a) We need to write down the sample space for two coin flips.

X={HH,HT,TH,TT}

P(HH)=0.25     X=2

P(HT)=0.25     X=1

P(TH)=0.25     X=1

P(TT)=0.25     X=0

Then

X=0, P(0)=0.25

X=1, P(1)=0.50

X=2, P(2)=0.25

b) If we filp the coin 4 times we have

4 H + 0 T   X=4    P(X)=1/16=0.0625

3 H + 1 T    X=3    P(X)=4/16=0.250

2 H + 2 T   X=2    P(X)=6/16=0.375

1 H + 3 T    X=1     P(X)=4/16=0.250

0 H + 4 T   X=0    P(X)=1/16=0.0625

c) If we flip the coin 10 times

10 H + 0 T         X=10     P(X)=1/1024=0.001

9 H + 1 T           X=9       P(X)=10/1024=0.0098

8 H + 2 T          X=8       P(X)=45/1024=0.0439

7 H + 3 T          X=7        P(X)=120/1024=0.1172

6 H + 4 T          X=6       P(X)=210/1024=0.2051

5 H + 5 T          X=5       P(X)=252/1024=0.2461

4 H + 6 T          X=4       P(X)=210/1024=0.2051

3 H + 7 T          X=3       P(X)=120/1024=0.1172

2 H + 8 T          X=2      P(X)=45/1024=0.0439

1 H + 9 T           X=1       P(X)=10/1024=0.0098

0 H + 10 T        X=0      P(X)=1/1024=0.001

"a" determine whether sage found the correct quotient. show your work

step-by-step explanation: