Abag contains chips of which 27.5 percent are blue. a random sample of 5 chips will be selected one at a time and with replacement. what are the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 ?

Step-by-step explanation:

We know that the mean and standard deviation of the sampling distribution of the sample proportion() is given by :-

, where p= Population proportion and n = sample size.

Let p be the proportion of blue chips.

As per given , we have

p= 0.275

n= 5

Now , the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 will be :

Hence, the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 are :

The standard deviation and mean is

Given that ;

No. of blue chips contain in a bag = P = 27.5 % = 0.27

No. of chips random selected at a time with replacement = n = 5

We know that the mean and standard deviation of the sampling distribution of the sample proportion  (p) is given by :-

= p

    = 0.275

=

Where p= Population proportion and n = sample size.

Let p be the proportion of blue chips.

As per given , we have

p= 0.275

n= 5

=

= 0.1997

Now , the mean and standard deviation of the sampling distribution of the sample proportion  of blue chips for samples of size 5 will be :

Hence, the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 are :

For more details about standard deviation and mean click the link given below;

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step-by-step explanation:

4y - 6 + 4y^2 - 9

before simplifying this contains 4 terms

after simplifying

4y -15 + 4y^2

it contains 3 terms