Question

Asked Jul 8, 2019

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See the attached question.

Step 1

**Introduction to binomial distribution:**

Consider a random experiment involving *n* independent trials, such that the outcome of each trial can be classified as either a “success” or a “failure”. The numerical value “1” is assigned to each success and “0” is assigned to each failure.

Moreover, the probability of getting a success in each trial, *p*, remains a constant for all the *n* trials. Denote the probability of failure as *q*. As success and failure are mutually exclusive, *q *= 1 – *p*.

Let the random variable *X* denote the number of successes obtained from the *n* trials. Thus, *X* can take any of the values 0,1,2,…,*n*.

Then, the probability distribution of *X* is a Binomial distribution with parameters (*n*, *p*) and the probability mass function (pmf) of *X*, that is, of a Binomial random variable, is given as:

Step 2

**Given information:**

Here, it is given that *X*(*n*,*p*) is a binomial random variable with parameters *n *and *p*.

Consider a binomial random variable *X*_{1}= *X*(*n*,*p*_{1}).

Consider another binomial random variable *X*_{2}= *X*(*n*,*p*_{2}).

If *X*(*n*,*p*) is a binomial random variable, then the mean and variance for the binomial variate *X *are *E*(*x*) = *n ** *p *and *V*(*x*) = *n ***p **(1 – *p*).

Since, *X*_{1} is a binomial random variable with parameters *n* and *p*_{1}, the mean of the binomial random variable *X*_{1 }is *E*(*X*_{1}) = *np*_{1}.

Since, *X*_{2} is a binomial random variable with parameters *n* and *p*_{2}, the mean of the binomial random variable *X*_{2 }is *E*(*X*_{2}) = *np*_{2}.

Step 3

**Calculation to show that X(n,p1) ≥st X(n,p2):**

From the below given calculation, it is proved that *X*(*n*,*p*1) ≥st *X*(*n*,*p*2):

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