A coin is tossed 100 times independently and lands heads 65 times. (a) Approxima

Question

A coin is tossed 100 times independently and lands heads 65 times. (a) Approximately, what is the probability of 65 or more heads if the coin is fair? (b) Do you think the coin is fair? Why?

6.5 Suppose (X,Y) has joint pdf f(x,y) = (1 + 2y)/4 for 0 x 2 and 0 y 1 and 0 otherwise.
(a) Are X and Y independent?

(b) find cov(X,Y).

(c) Use the marginal pdfs to find E(X) and E(Y).
(d) Use the joint pdf to find E(X + Y).
(e) Is E(X + Y) = E(X) + E(Y)? Did you expect this equality to hold? Why or why not? (d) Use the joint pdf to find E(XY).
(f) Is E(XY) = E(X)E(Y)? Did you expect this equality to hold? Why or why not?

In suitable units, X = time to observe a hazard and Y = time to react to it. (X,Y) has joint pdf f(x,y)=2for1x<y2andf(x,y)=0otherwise.Forparts(a)and(b),youcoulduseA dxdy=Area(A).

(a) Show that f(x,y) is a pdf.
(b) Compute P[X 1.4 and Y 1.6]. (c) Find the marginal pdf of X.
(d) Find the marginal pdf of Y.

Explanation / Answer

1) WE HAVE BEEN GIVEN THAT A COIN IS TOSSED 100 TIMES

AND HEAD COMES UP 65 TIMES

THEREFORE THE PROBABILITY OF HEAD = 0.65

PROBABILITY OF HEAD COMING 65 OR MORE = P(65)+P(66)+.....+P(99)+P(100)

WHERE EACH P(65) = 100C65*(0.65)^65*(0.35)^35 = 0.083

SIMILARLY WE HAVE TO CALCULATE TILL P(100)

AND THE ADDING ALL THE VALUES WE WILL GET THE REQUIRED ANSWER

THEREFORE P(65)+P(66)+...+P(99)+P(100) = 0.169

B) THE COIN IS A BIAS COIN BECAUSE IN A FAIR COIN THE PROBABILITY OF COMING HEAD = 0.50

BT AS PER THE GIVEN EXPERIMENT THE COIN LANDED UP 65 TIMES HEAD OUT OF 100 WHICH MAKES THE PROBABILITY OF COMING HEAD = 0.65

WHICH MEANS THAT THE COIN IS HIGHLY BIASED.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.