STAT 200 Week 3 Homework Problem PDF

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STAT 200 Week 3 Homework Problem University of Maryland University College June 7, 2019

4.1.4) The total of car is: 84+62+46+34+47+27=300 cars 

Let say A probability a person chose a car for safety.

P(A)= 84/300= 0.28 

Let say B probability a person chose a car for reliability.

P(B)= 62/300= 0.20 

Let say C probability a person chose a car coast.

P(C)= 46/300= 0.15 

Let say D probability a person chose a car for performance.

P(D)= 34/300= 0.11 

Let say E probability a person chose a car for comfort.

P(E)= 47/300= 0.16 

Let say F probability a person chose a car for looks.

P(F)= 27/300= 0.09

4.2.2) Number total of defects is: N= 5865+4613+1992+1838+1596+1546+1485+1398+1371+1130+1105+976+976 = 25891 a- Let say H probability of picking a lens that is scratched or flaked. P(H)= 5865+1992/ 25891= 0.30 b- Let say O probability of picking a lens that is the wrong PD or was lost in lab. 1

P(O)= 1398+976/ 25891= 0.09

c- Let say Z probability of picking a lens that is not scratched. Let say Z1 probability of picking a lens that is scratched. P(Z1)= 5865/25891= 0.23 P(Z)= 1 - P(Z1) P(Z)= 1 – 0.23 = 0.77

d- Let say X probability of picking a lens that is not the wrong shape. Let say X1 probability of picking a lens that is the wrong shape. P(X1)= 1485/ 25891 = 0.06 P( X)= 1- P(X1) = 0.94

4.2.8) a- Let say W probability of wining if you pick the number 7 and it comes up on the wheel. We have a space marked 0 to 36 and a space marked 00 so P(W) = 7/ 37 = 0.09

b- The odds against winning if you pick the number 7. Let say W2 probability of not wining if you pick the number 7 and it comes up on the wheel. P(W2) = 1 – P(W) = 1 – 0.09= 0.91 The odds against wining if you pick the number 7 is: Odds= P(W2) / P(W) = 0.91 / 0.09 = 10.11 The casino pays you $20 for every dollar you bet if your number comes up. It will pay you $10.11 for every $1 bet , on $20 they will pay : 10.11* 20= $202.22 2

Probability 0.8 0.7 0.6

The profit will be : $202.22 $20 = $182.11

0.5 0.4 0.3 0.2

4.4.6)

0.1 Probability

0

10P6

= 10! / (10-6) !

Number of hesds

= 10! / 4! = 10*9*8*7*6*5*4*3*2*1 / 4*3*2*1 10P6 = 151200

4.4.12) In this case the order doesn’t matter so it a combination of 20 people out of 7 people. 20C7

= 20! / 7! (20-7)! = 20! / 7! 13!

20C7

= 77520

5.1.2 a- The state of the variable S= ( HHH , HTH , TTT , THT) b- Let say B probability distribution for the number of heads. P(B)= 3/4 = 0.75 c- Histogram

3

d- The mean Mean= x p(x) = 3 * 0.75 Mean= 2.25 e- The variance Variance= (x-u)2 P(x) = (3 – 2.25)2 * 0.75 Variance = 0.42

The standard deviation Standard deviation = √(x-u)2 P(x) = √0.42 Standard deviation = 0.65

g- probability of having two or more number of heads.

Let say M probability of having two or more number of heads P(M) = 2/4 = ½ = 0.5 h- Since the probability of having two or more number of heads is 50% we can say that it’s unusual to flip two heads.

5.2.4) a- P(x=1) = 6C1 P1 q6-1 = 6C1 (0.3) * (1-0.3)5 P(x=1) = 0.302 b- P(x=5) = 6C5 (0.3)5 *( 1-0.3)1 = 0.01 c- P(x=3)= 6C5 (0.3)3* ( 1-0.3)3 = 0.05 4

P(x≤ 3) = P(x=1) + P(x=2) + P(x=3) = 0.0302+ 0.324+0.05 = 0.676

f- P(x≤4)= P(x=1) + P(x=2)+ P(x=3)+P(x=4) = 0.320+0.324+0.05+0.06 = 0.75 e- P(x≥5)= 1- P(x≤4) = 1- 0.75 = 0.25

5.2.10) a- The random variable is : n= 52 , P= 14%= 0.14 , q= 1-0.14= 0.86

b- This is a binominal experience because the n=52 is independent , the result will not influence the outcomes of other trials.

c- P(x=6) = 52C6 (0.14)6 * (0.86)46 = 0.15 d- P(x=25)= 52C25(0.14)25* (0.86)27 = 3.66 10-9

e- P(x=52)= 52C52 (0.14)52 * (0.86)0 = 3.97 10-45 f- This can’t be unusual for package to have only brown M&M’

5.3.4) a- The random variable: 5

n= 50 , P= 10%= 0.10 , q= 1-0.10= 0.9

b- The probability distribution is P=0.10

c-

Probability 0.12 0.1 0.08 0.06 0.04 0.02 0

Distribution Probability

From the chart above we notice that the probability of all people who are left-handed is low among the 15 people.

e- The median: u= xP(X)= 15*0.1= 1.5 u= 1.5 f- The variance: Variance= (x-u)2 P(x) =( 15-1.5)2 *0.1 = 18.22 g- Standard deviation: Standard deviation= √18.22 = 4.27

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