RESPOND TO THESE COMMENTS BY EXPRESSING OPINION AND SUGGESTION.

1)The number I chose for my extended family was 15 because that seems roughly about how many family members I have. The attribute I chose was how many members have a pet. When I looked it up it said that roughly 70% of people have a pet at home. The mean for this group is 8 and the standard deviation is 4.47. After plugging in these statistics to the normal calculator on stat crunch it then shows my binomial distribution. When I press the between option and plug in my 0.70 probability this concludes that between 3 and 13 people have a pet which is 10/15 which is roughly 70%. Also to test this theory when I plug in these statistics to the standard option I also get 10/15. When I make up my own scenario of (6

2)For this discussion assignment, I decided to look at the attribute of left-handedness in my extended family of 25 (n=25) people. After doing some internet research I discovered that the probability of being left-handed is roughly 10-15% according to medlineplus.gov, so I chose to go with .12 as the probability in my experiment (p= .12).

This is a great attribute for a binomial probability experiment because it meets all 4 of the criteria. The first criteria is that the experiment is performed a fixed number of times, in this case there are 25 trials. The second criteria is that all trials are independent. The outcome of one person's left or right-handedness will not affect the outcome of the next. The third criteria is that there are two mutually exclusive outcomes. In this experiment, these two outcomes are either right-handedness or left-handedness. The 4th and final criteria is that the probability of success is the same for each trial. The probability will remain at .12 for each of the 25 trials.

The mean of my distribution is 3. The standard deviation of my distribution is 1.62.

I played around with some sample probabilities. In order toperform this, I used the following command:

Statcrunch>calculator>binomial>standardorbetween>compute

First, I found outP(X4), or the probability that there were at least 4 family members who would be left-handed. The probability was .35. Then, I figured outP(X<5), or the probability that the number of people in my family who are left-handed was less than 5. The probability was .83. Lastly, I wanted to see what the probability would be that at least 2 people were left-handed, but no more than 8 people, orP(2X8).The resulting probability of this equation was .82. This is what that distribution for that equation looked like:

I can use the approximation to the normal distribution with my binomial because mynp(1p)10. This is calculated with n= 25 and p= .12 for my distribution.