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Distribution

Chapter 5

5.32 Expected values of a probability distribution are defined as the weighted mean of possible values. After many trials in an experiment, the outcome is the expected value.

5.33 Properties of Binomial Distribution

  • There should be a fixed number of trials
  • Trials should be independent
  • There should be two possible outcomes; success and failure
  • The probability of success and failure does not change from one trial to another

5.34 properties of Poison Distribution

  • Outcomes should be classified as success or failure
  • Number of success should be known
  • Probability of success is equivalent to the size of the region
  • Probability of success approaches zero as the size of the region becomes smaller

5.38 (a) To find the probability that Dow Jones Industrial Average increases in 9 or more of the 12 U.S presidential elections, we use binomial distribution.

P=0.5, q=0.5, n=12, and x=9, 10, 11, 12

Thus, P(X)= ((n!/(n-x)!x!))*p^x*q^(n-x)

P(X9)=

((12!/(12-9)!9!))*0.5^9*0.5^3= 0.0537

P(X10)= ((12!/(12-10)!10!))*0.5^10*0.5^2= 0.0161

P(X11)= ((12!/(12-11)!11!))*0.5^11*0.5^1= 0.0029

(PX12)= ((12!/(12-12)!12!))*0.5^12*0.5^0= 0.0002

= 0.0729

(b) As defined in the problem, there is 0.75 probability of success. This means that the probability of failure will be 0.25.

P= 0.75, q= 0.25, n=12, and x=9, 10, 11, 12

P(X9)=

((12!/(12-9)!9!))*0.75^9*0.25^3= 0.2581

P(X10)= ((12!/(12-10)!10!))*0.75^10*0.25^2= 0.2323

P(X11)= ((12!/(12-11)!11!))*0.75^11*0.25^1= 0.1267

(PX12)= ((12!/(12-12)!12!))*0.75^12*0.25^0= 0.0317

= 0.6488

5.39 we use binomial distribution to calculate the probability of bills having errors.

n=10, p=0.4, q=0.6

  1. P(X=0)= ((10!/(10-0)!0!))*0.4^0*0.6^10= 0.00604
  2. P(X=1)= ((10!/(10-1)!1!))*0.4^1*0.6^9= 0.0403

5.45

  1. a) We use binomial distribution where n=44, x=35, 36,…,44, p=0.5, and q=0.5

P(X=35)= ((44!/(44-35)!35!)*0.5^35*0.5^9= 0.00004029

P(X=36)= ((44!/(44-36)!36!)*0.5^36*0.5^8= 0.00001007

P(X=37)= ((44!/(44-37)!37!)*0.5^37*0.5^7= 0.0000021

P(X=38)= ((44!/(44-38)!38!)*0.5^38*0.5^6= 0.0000001

P(X=39)= ((44!/(44-39)!39!)*0.5^39*0.5^5= 0.0000001

P(X=40)= ((44!/(44-40)!40!)*0.5^40*0.5^4= 0.0000001

P(X=41)= ((44!/(44-41)!41!)*0.5^41*0.5^3=0.0000001

P(X=42)= ((44!/(44-40)!42!)*0.5^42*0.5^2=0.0000001

P(X=43)= ((44!/(44-43)!43!)*0.5^43*0.5^1=0.0000001

P(X=44)= ((44!/(44-44)!44!)*0.5^44*0.5^0= 0.0000001

Total probability= 0.000053

  1. b) As the number of success approach the total number of trials, the probability approaches zero. Thus, the probability that the indicator would be correct diminishes as the number of success increase.

Chapter 6

6.23 The normal table is used because many variables in statistics exhibit a normal probability distribution. The graph of many variables used in statistics is often a normal curve.

6.24 The first step involves calculating the z-scores corresponding to the two values. Secondly, you find the corresponding probabilities from the normal table. Finally, you find the difference between the large value and the smaller value. The results of the subtraction represent the area between the two points.

6.25 since the formula for finding the percentiles in the normal distribution is given by z=(x-µ)/δ, finding x will require rearranging the formula. The rearranged formula will be given as X= (z*δ)+µ, where z represents the percentile, δ represents the standard deviation, and µ represents the mean of the normal distribution.

6.26 The properties of a normal distribution are:

  • The mean, mode, and median are the same
  • It is perfectly symmetrical around the centre
  • 50% of data lie below the mean and another 50% lie above the mean

6.27 The shape of the normal distribution differs from the uniform and exponential distribution shapes in that, while the uniform distribution has a piecewise constant density, the normal curve is bell-shaped. On the other side, the exponential distribution curve is one sided-curve.

6.28 The process of determining if a set of data is normally distributed involves drawing a graph of z-scores against the data. The process requires one to determine the corresponding z-scores of every value in the data. If the resulting line graph is diagonal and straight, the data is normally distributed.

 

 

 

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