can you help solve this statistic questions? thanks very much

We are given the incidences of the number of broken mugs in 100 different containers.

(a) The mean, or expected value, is found by taking the sum of the product of each occurence by the frequency and dividing by the total. (Equivalently, we can calculate the individual probabilities and convert the table to a probability distribution. In this case this is very easy as the total is 100, so the probability of 0 broken is...

We are given the incidences of the number of broken mugs in 100 different containers.

(a) The mean, or expected value, is found by taking the sum of the product of each occurence by the frequency and dividing by the total. (Equivalently, we can calculate the individual probabilities and convert the table to a probability distribution. In this case this is very easy as the total is 100, so the probability of 0 broken is 41/100=.41, etc... Then the mean is the sum of the products of the variable and the probabilitites.)

mu=1(.41)+2(.25)+3(.07)+4(.02)+5(0)+6(0)=1.2

or [1(41)+2(25)+3(7)+4(2)+5(0)+6(0)]/100=1.2

(b) The requirements for a binomial distribution are:

(1) There are a fixed number of trials.
(2) Each trial can have only two outcomes (or be reduced to two outcomes.)
(3) The outcomes should be independent.
(4) The probability of success should remain constant.

We can model this with a binomial distribution since there are a finite number of trials (in this case 6 categories), the outcomes are broken/ not broken, the outcomes are independent (cannot have 1 broken and 2 broken), and the probability does not change.

The mean or expected value of a binomial distribution is calculated by mu=np where n is the number of outcomes and p the probability of each. We calculated mu=1.2, and n=6 so 1.2=6p ==> p=.2

(c) We can calculate the probability for each outcome by:

P(X)=C(n,x)(p)^x(1-p)^(n-x) where C(n,x) is the number of combinations of n objects taken x at a time.

Using an algebra utility we find:

X          0         1         2         3          4         5          6
P(x)     .262    .393    .246     .082     .015     .002     .00006
Expected 26     39        25       8          2         0          0
Observed 25     41        25       7          2         0          0

The distribution appears to be a good estimate.