You are told that 16.6% of dog breeders have more than 2 years of experience bre

Question

You are told that 16.6% of dog breeders have more than 2 years of experience breeding bulldogs, and 6.68% of breeders have less than 9 month of experience breeding bulldogs. If this data assumes normal distribution, what is the value of the mean and standard deviation of this distribution?

A study was conducted to determine how often bulldogs have litters that contain more male than female puppies. It was determined that 59% of the time litters contain more males than females. Using the normal approximation to the binomial, what is the probability that out of a sample of 70 bulldog litters, fewer than 35 litters have more males than female puppies?

Explanation / Answer

Q1)As We know that the data is normally distributed we can use the following Normal Distribution formula

Z=(X'-mu)/sd so.,

sd*Z=X'-mu so.,

mu+sd*Z=X' --(A)

Now the Z value at (1-0.166) is 0.9700 (From z table) & X'=2 is given

so putting these values in A the equation becomes

mu+0.97*sd=2 (i)

Same way

Now the Z value at 0.0668 is -1.5 (From z table) & X'=9/12 is given

so putting these values in A the equation becomes

mu-1.5sd=9/12 ----(ii)

solving equation (i) & (ii) we get

2.47sd=1.25

sd=0.5060729 Year &

mean=1.509109 Year

Q2) Now it's binomial distribution sum

P(X=x)=nCx * P^x * (1-P)^(n=x)

we have p=0.59

n=70

x<35

P(X<35)=P(X=0)+P(X=1)+....+P(X=34) = 0.05013635 (You have to use some software to solve this as you can not put all the values shown above and solve the same for 35 times (i.e. 0 to 34) . I have done the same in R and solved the same with the below formula. Just FYI the same is provided below. You can use any tool for the same:

>sum(dbinom(0:34,70,.59))
[1] 0.05013635

Hope the above explaination has helped you in understanding the problem. Please upvote if it has really helped. Good Luck!!!

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