Mathematics, 18.02.2020 00:35 desiwill01

The heights, in inches, of the starting five players on a college basketball team are 6868, 7373, 7777, 7575, and 8484. Considering the players as a sample, the mean and standard deviation of the heights are 75.475.4 inches and 5.95.9 inches, respectively. When the players are regarded as a population, the mean and standard deviation of the heights are 75.475.4 inches and 5.25.2 inches, respectively. Explain why, numerically, the sample mean of 75.475.4 inches is the same as the population mean but the sample standard deviation of 5.95.9 inches differs from the population standard deviation of 5.25.2 inches

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The sample standard deviation of 5.95.9 inches differs from the population standard deviation of 5.25.2 inches because of their formulas for calculating it.

Step-by-step explanation:

We are given the heights, in inches, of the starting five players on a college basketball team ;

68, 73, 77, 75 and 84

Now whether we treat this data as sample data or population data, the mean height would remain same in both case because the formula for calculating mean is given by ;

Mean = Sum of all data values ÷ No. of observations

Mean = ( 68 + 73 + 77 + 75 + 84 ) ÷ 5 = 75.4 inches

So, numerically, the sample mean of 75.4 inches is the same as the population mean.

Now, coming to standard deviation there will be difference in both sample and population standard deviation and that difference occurs due to their formulas;

Formula for sample standard deviation = $\frac{\sum (X_i - Xbar)^{2} }{n-1}$

where, X_i = each data value

X bar = Mean of data

n = no. of observations

Sample standard deviation = $\frac{ (68 - 75.4)^{2} +(73 - 75.4)^{2}+(77- 75.4)^{2}+(75- 75.4)^{2}+(84 - 75.4)^{2} }{5-1}$

= 5.9 inches

Whereas, Population standard deviation = $\frac{\sum (X_i - Xbar)^{2} }{n}$

= $\frac{ (68 - 75.4)^{2} +(73 - 75.4)^{2}+(77- 75.4)^{2}+(75- 75.4)^{2}+(84 - 75.4)^{2} }{5}$ = 5.2 inches .

So, that's why sample standard deviation of 5.95.9 inches differs from the population standard deviation of 5.25.2 inches only because of formula.

Answer from: Quest

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

$x^2-2x+y^2-6y=26$

we now create space between the x and y terms with parenthesis.

$(x^2-2x)+(y^2-6y)=26$

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

$\frac{-2}{2} =(-1)^{2} =1$

$\frac{-6}{2} =(-3)^{2} =9$

we add the squares to both sides.

$(x^2-2x+1)+(y^2-6y+9)=26+1+9$

simplify.

$(x^2-2x+1)+(y^2-6y+9)=36$

and write the quadratics in factored form.

$(x-1)^{2} +(y-3)^{2} =36$

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.

Answer from: Quest

the answer for this question is a= 13,622.84

step-by-step explanation:

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