The random variable K has a geometric distribution with mean 16. Which of the following is closest to the standard deviation of random variable K ?

a. 0.0625
b. 0.9375
c. 4
d. 15.49
e. 240

Respuesta :

Answer: d. 15.49

Step-by-step explanation:

Given : The random variable K has a geometric distribution with mean 16.

Formula for Mean and standard deviation in geometric distribution:

[tex]\mu=\dfrac{1}{p}[/tex]    (1)

[tex]\sigma=\dfrac{\sqrt{1-p}}{p}[/tex]   (2), where p is the probability of success in each trial.

Since [tex]\mu=16[/tex]  (Given)

Then from (1) ,

[tex]\dfrac{1}{p}=16\\\Rightarrow\ p=\dfrac{1}{16}=0.0625[/tex]

Put value of p in (2),

[tex]\sigma=\dfrac{\sqrt{1-0.0625}}{0.0625}=\dfrac{\sqrt{0.9375}}{0.0625}[/tex]  

[tex]=\dfrac{0.968245836552}{0.0625}[/tex]

[tex]=15.4919333848\approx15.49[/tex]

Hence, the standard deviation of random variable K is closest to 15.49.

Therefore , the correct answer is d. 15.49 .

The standard deviation of random variable K is 15.49 and this can be determined by using the formula of mean and standard deviation.

Given :

The random variable K has a geometric distribution with a mean 16.

The mean is given by the formula:

[tex]\mu = \dfrac{1}{p}[/tex]

Substitute the value of [tex]\mu[/tex] in the above formula in order to determine the value of 'p'.

[tex]16 = \dfrac{1}{p}[/tex]

p = 0.0625

The standard deviation is given by the formula:

[tex]\rm \sigma =\dfrac{ \sqrt{{1-p}}}{p}[/tex]

Now, substitute the value of p in the above formula in order to determine the standard deviation.

[tex]\rm \sigma =\dfrac{ \sqrt{1-0.0625}}{0.0625}[/tex]

[tex]\rm \sigma = \dfrac{\sqrt{0.9375} }{0.0625}[/tex]

[tex]\sigma = 15.49[/tex]

Therefore, the correct option is d).

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https://brainly.com/question/17921485