Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,979. The standard deviation of the sample was $1,000.
a. Based on this sample information, develop a 90 percent confidence interval for the population mean yearly premium.
b. How large a sample is needed to find the population mean within $250 at 99 percent confidence?

Given Information:

Standard deviation = σ = $1000

Mean = μ = $10,979

Confidence level = 90%

Margin of error = $250

Sample size = n = 20

Required Information:

a) 90%  confidence interval = ?

b) Sample size at 99% confidence = n = ?

a) 90%  confidence interval =  ($10,611.17, $11,346.83)

b) Sample size at 99% confidence = n = 107

Step-by-step explanation:

a) The confidence interval is given by

confidence interval = μ ± z*σ/√n

Where μ is the mean annual health insurance premium, σ is the standard deviation, n is number of subscribers that were surveyed, and z is the corresponding z-score for 90% confidence level.

The z-score for 90% confidence level is 1.645

Confidence interval = 10,979 ± (1.645*1000)/√20

Confidence interval = 10,979 ± 367.83

upper limit = 10,979 + 367.83 = 11,346.83

lower limit = 10,979 - 367.83 = 10,611.17

Confidence interval = ($10,611.17, $11,346.83)

Which means that we are 90% confident that the annual health insurance premiums for a family with coverage through an employer is between  ($10,611.17, $11,346.83)

b) To find the required sample size

Me = z*(σ/√n)

√n = z*σ/Me

n = (z*σ/Me)²

Where z is the corresponding z-score, n is the required sample size σ is the standard deviation and Me is the margin of error that is $250

The z-score for 99% confidence level is 2.576

n = (2.576*1000/250)²

n = 106.17

n = 107

Therefore, a sample size of 107 is needed to find the population mean within $250 at 99% confidence.

perimeter of the first rectangular board is 9 feet.

step-by-step explanation:

let p = perimeter of first board.

l = length of first board

w = width of first board

length of first board = 2 * width of first board - 3

l = 2w - 3

perimeter of first board = 2l + 2w

since l = 2w-3, then

p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6

length and width of the second board are reciprocals of corresponding length and width of the first board.

length of second board = 1/(2w-3)

width of second board = 1/w

perimeter of second board is 1/5th perimeter of the first board.

perimeter of second board = p/5

perimeter of the second board = (2/(2w-3) + 2/w)

since the perimeter of the first board is 5 times the perimeter of the second board, this means that:

6w-6 = 5*(2/(2w-3) + 2/w)

which is the same as:

6w-6 = (10/(2w-3) + 10/w)

if we multiply both sides of this equation by (2w-3), we get:

(6w-6)*(2w-3) = 10 + 10*(2w-3)/w

if we multiply both sides of this equation by w, we get:

(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)

we can simplify this to become:

(6w-6)*(2w-3)*w = 10w + 20w - 30

which becomes:

(6w-6)*(2w-3)*w = 30w-30

if we divide both sides of this equation by (6w-6), we get:

(2w-3)*w = (30w-30)/(6w-6)

which becomes:

(2w-3)*w = 5

this can be simplified to:

2w^2 - 3w = 5

subtract 5 from both sides of this equation to get:

2w^2 - 3w - 5 = 0

which can be factored into:

(2w-5)*(w+1) = 0

which makes:

w = 5/2

or

w = -1.

w can't be negative so the only possible answer is w = 5/2.

if w = 5/2, then

p = 2w-3 = 5-3 = 2

we have:

p = 2

w = 5/2

p = 2l + 2w gets p = 4 + 5 = 9

the perimeter of the first board is 9 feet.

since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.

to see if that's correct, we substitute known values for l and w into the second board.

the length of the second board is 1/l.

this becomes 1/2.

the width of the second board is 1/w

this bgecomes 1/(5/2) = (2/5)

let l2 = length of second board.

let w2 = width of second board.

l2 = (1/2)

w2 = (2/5)

let p2= perimeter of second board.

then p2 = 9/5

since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.

2 * l2 = 2 * (1/2) = 1

2 * w2 = 2 * (2/5) = 4/5

1 is the same as 5/5

5/5 + 4/5 = 9/5

p2 checks out good.

the dimensions as given are good.

answer to the problem is:

perimeter of the first rectangular board is 9 feet.