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StatisticsQ&A Library3. According to a 2015 study, 33% (033) of US, households have at least one ultra high-definition television. Use the Normal Approximation to the Binomial to calculate theprobabilitdefinition television.y that 4 or fewer households out of a sample of 16 will have at least one ultra high-Question

Asked Mar 29, 2019

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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

**Parameters of binomial distribution for the given situation:**

The size of the sample taken from the population of U.S households is *n* = 16. It is known that each and every household will not depend on the other, imp-lying independence among each other. As a result, the 16 households may be considered as 16 independent trials.

Consider the event that house hold having at least one ultra high-definition television as a “success”. It is given that 33% of U.S households have at least one ultra high-definition television. Thus, the probability that house hold having at least one ultra high-definition television, that is, the probability of success in each trial is *p* = 0.33. Then *q* = 1 – 0.33 = 0.67.

Consider *x* as the number of house hold having at least one ultra high-definition television among a sample of 16 U.S households. Then, *x* has a Binomial distribution with parameters (*n* = 16, *p* = 0.33) with pdf as given below:

*P*(*x*) = ^{16}*C _{x}**(0.33)

Step 3

**Normal approximation to binomial distribution:**

The mean and variance for the binomial variate is *E*(*x*) = *n ** *p *and *V*(*x*) = *n ***p **(1 – *p*).

Here, *p *= 0.33, 1 – *p *= 0.67 and *n *= 16.

The mean and variance of the variable *x *(number of house hold having at least one ultra high-definition television) is *E*(*x*) = *n ** *p *= 16*0.33 = 5.28 and *V*(*x*)...

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