Question 1- Binomial Probability Distribution If you are given that a manufactur

Question

Question 1- Binomial Probability Distribution

If you are given that a manufacturing process follows a binomial distribution with the probability of defects equal to 0.05. Consider that you randomly select 50 objects.

a. What are the mean, variance, and standard deviation for the process?

b. What is the probability of seeing exactly 1 defect? c. What is the probability of seeing 2 or more defects in the 50 objects? (ie P(X 2)) hint: use the complemental rule here.

2 Question 2 – Normal Probability Distribution Suppose you are told that average IQ of a group of people follow a normal probability distribution with a mean of 110 and standard deviation of 7.5.

a. What is the probability that a randomly selected person will have a value of 105 or greater?

b. What percentage of people will have IQ scores between 105 and 135?

Explanation / Answer

Q1

Binomial distributions are functions of "n" number of sample and "p" Probability of specific occurances . Hence Binomial Distributions are written as B(n,P) = nCx* (P)^(x) * (1-P)^(n-x)

Mean for binomial distribution is given as equals to =n*P

Variance equals to = n*P*(1-P)

Standard Deviation = Square root (n*P*(1-P)) ; Here n = 50 & P = 0.05

Hence Mean = 50*0.05 = 2.5

Variance = 50*0.05(1-0.05) = 2.5*(0.95) = 2.375

SD = Sq Root( Variance) = Sq Root (2.375) = 1.543

P(x=1) = 50C1* (0.05)^1*(0.95)^49 = 0.202

We know that summation of all point probabilities over a scale of probailites always equals to 1

Here in this case; P(x<2) + P(x>=2) equals to 1

P(x>=2) = 1-P(x<2)

We have P(x<2) =P(x=0) + P(x=1)

P(x=0) = 50C0 * (0.05)^0* (0.95)^50 = 0.076

P(x<2) = 0.076 + 0.202 = 0.278

P(x>=2) = 1-P(x<2) = 1-0.278 = 0.922

Q2.

For Normal distribution we use Z test for large samples generally more than 30

Z = (x-mean)/standard deviation for any Normal distribution N(mean,SD)

hence x= mean+ Z*Standard Deviation

in this case x =105

105 = 110 + z*7.5 => z =5/7.5 =0.66

Area to the left of x = 105 is (z=0.66) = 0.24

Hence if x>105 then Probability should be 1-0.24 =0.76

Q.b

for x = 135 we have z =135-110/7.5 =25/7.5 =3.33

and z value for 3.33 is 0.99

hence 99% area is covered when x = 135 & 24% of area covered when x=105 therefore 99%-24% =75% of people will have IQ scores between 105 & 135.

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