Question

It is known that 80% of all brand A external hard drives work in a satisfactory manner throughout the warranty period (are "successes"). Suppose that n= 15 drives are randomly selected. Let X = the number of successes in the sample. The statistic X/n is the sample proportion (fraction) of successes. Obtain the sampling distribution of this statistic.

Answer 1

Answer:

P (x= 5) =  0.0001

P(x=3) =  0.008699

Step-by-step explanation:

This is a binomial distribution .

Here p = 0.8  q= 1-p = 1-0.8 = 0.2

n= 15

So we find the probability for x taking different values from 0 - 15.

The formula used will be

n Cx p^x q^n-x

Suppose we want  to find the value of x= 5

P (x= 5) = 15C5*(0.2)^10*(0.8)^5 = 0.0001

P(x=3) = 15C3*(0.2)^12*(0.8)^3 =  9.54 e ^-7= 0.008699

Similarly we can find the values for all the trials from 0 -15  by substituting the values of x =0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15.

Answer 2

The value for p(x = 5) is 0.0001 and the value for p(x = 3) is 0.008699.

It is given that the 80% of all brand A external hard drives work in a satisfactory manner throughout the warranty period.

It is required to find the sampling distribution if n =15 samples.

What is sampling distribution?

It is defined as the probability distribution for the definite sample size the sample is the random data.

We have p =80% = 0.8 and q = 1 - p ⇒ 1 -0.8 ⇒ 0.2

n = 15

We can find the probability for the given x by taking different values from 0 to 15

the formula can be used:

$\rm _{n}^{}\textrm{C}_x p^xq^{n-x}$

If we find the value for p(x = 5)

$\rm _{15}^{}\textrm{C}_5 p^5q^{15-5}\\\\\rm _{15}^{}\textrm{C}_5 0.8^50.2^{10}$⇒ 0.0001

If we find the value for p(x = 3)

$\rm _{15}^{}\textrm{C}_3 0.8^30.2^{12}$ ⇒  

Similarly, we can find the values for all the trials from 0 to 15 by putting the values of x = 0 to 15.

Thus, the value for p(x = 5) is 0.0001 and the value for p(x = 3) is 0.008699.

Learn more about the sampling distribution here:

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