Question

1. In the game of​ roulette, a player can place a $7 bet on the number 15 and have a 1/38 probability of winning. If the metal ball lands on 15, the player gets to keep the ​$7

paid to play the game and the player is awarded ​$245. Otherwise, the player is awarded nothing and the casino takes the​ player's $7. What is the expected value of the game to the​ player? If you played the game 1000​ times, how much would you expect to​ lose?

The expected value is

​$_______​(Round to the nearest cent as​ needed.)

The player would expect to lose about

​$_______

2.

Use

n equals=6

and

pequals=0.35

to complete parts​ (a) through​ (d) below.

​(a) Construct a binomial probability distribution with the given parameters.

x	​P(x)
0	nothing
1	nothing
2	nothing
3	nothing
4	nothing
5	nothing
6	nothing

​(Round to four decimal places as​ needed.)

​(b) Compute the mean and standard deviation of the random variable using

mu Subscript Upper XμXequals=Summation from nothing to nothing left bracket x times Upper P left parenthesis x right parenthesis right bracket∑[x•P(x)]

and

sigma Subscript Upper XσXequals=StartRoot Summation from nothing to nothing left bracket x squared times Upper P left parenthesis x right parenthesis right bracket minus mu Subscript Upper X Superscript 2 EndRoot∑x2•P(x)−μ2X.

mu Subscript Upper XμXequals=nothing

​(Round to two decimal places as​ needed.)

sigma Subscript Upper XσXequals=nothing

​(Round to two decimal places as​ needed.)​(c) Compute the mean and standard​ deviation, using

mu Subscript Upper XμXequals=np

and

sigma Subscript Upper XσXequals=StartRoot np left parenthesis 1 minus p right parenthesis EndRootnp(1−p).

mu Subscript Upper XμXequals=nothing

​(Round to two decimal places as​ needed.)

sigma Subscript Upper XσXequals=nothing

​(Round to two decimal places as​ needed.)

​(d) Draw a graph of the probability distribution and comment on its shape.

Which graph below shows the probability​ distribution?

A.

03600.250.5xP(x)

A graph with a horizontal x-axis labeled from 0 to 6 in intervals of 1 and a vertical y-axis labeled from 0 to 0.5 in intervals of 0.05 has seven vertical line segments, one for each horizontal axis label. The approximate heights of the vertical line segments are as follows, with the horizontal coordinate listed first and the approximate line segment height listed second: 0, 0.08; 1, 0.24; 2, 0.33; 3, 0.24; 4, 0.10; 5, 0.02; 6, 0.00.

B.

03600.250.5xP(x)

A graph with a horizontal x-axis labeled from 0 to 6 in intervals of 1 and a vertical y-axis labeled from 0 to 0.5 in intervals of 0.05 has seven vertical line segments, one for each horizontal axis label. The approximate heights of the vertical line segments are as follows, with the horizontal coordinate listed first and the approximate line segment height listed second: 0, 0.08; 1, 0.24; 2, 0.33; 3, 0.24; 4, 0.10; 5, 0.24; 6, 0.33.

C.

03600.250.5xP(x)

A graph with a horizontal x-axis labeled from 0 to 6 in intervals of 1 and a vertical y-axis labeled from 0 to 0.5 in intervals of 0.05 has seven vertical line segments, one for each horizontal axis label. The approximate heights of the vertical line segments are as follows, with the horizontal coordinate listed first and the approximate line segment height listed second: 0, 0.24; 1, 0.10; 2, 0.02; 3, 0.00; 4, 0.02; 5, 0.10; 6, 0.24.

D.

03600.250.5xP(x)

A graph with a horizontal x-axis labeled from 0 to 6 in intervals of 1 and a vertical y-axis labeled from 0 to 0.5 in intervals of 0.05 has seven vertical line segments, one for each horizontal axis label. The approximate heights of the vertical line segments are as follows, with the horizontal coordinate listed first and the approximate line segment height listed second: 0, 0.10; 1, 0.24; 2, 0.33; 3, 0.33; 4, 0.33; 5, 0.24; 6, 0.10.The binomial probability distribution is

▼

Answer 1

1)

expected value is =(245)*(1/38)-7*(37/38)=-0.37

If you played the game 1000​ times, player would expect to lose about =$368.42

2)

x	f(x)
0	0.0754
1	0.2437
2	0.3280
3	0.2355
4	0.0951
5	0.0205
6	0.0018

E(x) =μ=

ΣxP(x) =

2.1000

std deviation=

σ= √σ2 =

1.17

c)

E(x) =μ=

np=6*0.35=

2.1000

std deviation=

σ =

1.17

d)