1 Consider the following experiment Three fair coins are fli

1. Consider the following experiment: Three fair coins are flipped one after the other attaching a value of 1 to tails and a value of 2 for heads. The values from all three flips are multiplied together for a final value X(for example, if the result is heads, tails, heads, one would multiply 2 . 1 . 2 so X = 4) (a) Is X a discrete or continuous random variable? (b) What is the probability distribution for X? Express this distribution in a table. (c) What is the expected value, or population mean E(X)? (d) What is the variance of X? 2. Consider the following data sampled from students on hours slept the night before the midterm exam and performance on that exam: Student Hours Slept Test Score 60% 90% 80% 95% en Vincent Michael Angela (a) Draw a scatter plot of the data below with Test Score on the Y-axis and Hours Slept on the X-axis. Do Test Score and Hours Slept look positively or negatively correlated? (b) What is the sample mean of Test Score? (c) What is the sample mean of Hours Slept? (d) What is the sample correlation coefficient between Test Score and Hours Slept?

Solution

Ans 1)

Yes this is discrete random variable because values taken by X are either 1 or 2. In case of continuous variable it can take any real number i.e 0.00001 , 0.00000023 etc.

Ans 2)

Probability distribution of X is depends upon the outcome generated from 3 coins i.e either head or tail

Number of Heads Probability

0 1/8 ( Only one such scenario is possible

1   3/8 (Either of the coin has it hence 3 times out of sample sized 8)

2 3/8 ( Only one coin has tail and that can be any of these 3 coins)

3 1/8 ( Only once it is possible)

Sample Size is 8 because (H,H,H),(T,T,T),(H,T,T),(T,H,T),(T,T,H),(H,H,T),(H,T,H),(T,H,H)

Ans 1/3)

Expected Valeu of X=sumof (Probability of outvome* Possible outcome) from above sample set we can calculate the value of each set for an example (H,H,H)=2+2+2=6 (T,H,T)=1+2+1=4

6,3,4,4,4,5,5,5 hence E(X)=(1/8)*6+(1/8)*3+(3/8)*4+(3/8)*5=36/8=4.5

Ans 1/4)

Variance (X)=E(X^2)-(E(X)^2)

E(X^2)=(1/8)*6^2+(1/8)*3^2+(3/8)*4^2+(3/8)*5^2=21

Var(X)=21-(4.5^2)=21-20.25=0.25