A public health official is planning for the supplyof influenza vaccine needed for the upcoming flu season. She wants to estimate the proportion p who plan to get the vaccine.

How large does the sample size n need to be for X/n, the sample proportion of people who plan to get the vaccine, to have an 90% chance of lying within 0.04 of probability?

And rounded up we have that n=421

Step-by-step explanation:

We know that the sample proportion have the following distribution:

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by and . And the critical value would be given by:

The margin of error for the proportion interval is given by this formula:  

   (a)  

And on this case we have that and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

  (b)  

We assume that a prior estimation for p would be since we don't have any other info provided. And replacing into equation (b) the values from part a we got:

And rounded up we have that n=421

d) 30 units^2

area of rectangle = l × b

we know,

length = 10 unit

breadth = 3 unit

applying the formula,

= l × b

= 10 × 3

= 30 unit^2

happy to

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ending amount = beginning amount / (2^n)

where "n" = number of half-lives (or elapsed time / half-life)

ending amount = 50 mg / (2 ^10 / 5.3)

ending amount = 50 mg / (2 ^ 1.8867924528)

ending amount = 50 m / 3.6981210597

ending amount = 13.5203794557 mg

or about 13.5 mg remain

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