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trapecia [35]
3 years ago
12

The following table gives the scores of 30 students in a mathematics examination. Scores 90–99 80–89 70–79 60–69 50–59 Students

3 7 12 4 4 Find the mean and the standard deviation of the given data. Hint: Assume that all scores lying within a group interval take the middle value of that group. (Round your answers to two decimal places.)
Mathematics
1 answer:
Pavel [41]3 years ago
6 0
The given data is
Scores:      90-99  80-89  70-79  60-69  50-59
Students:   3          7           12        4          4
Because we are to take the middle value for each group.
Therefore the total score is
Sx = 3*94.5 + 7*84.5 + 12*74.5 + 4*64.5 + 4*54.5 = 2245
The number of students is 3+7+12+4+4 = 30

The mean is
m =2245/30 = 74.833

Calculate (Sx - m)² 
(Sx - m)² = 3*(94.5-74.833)² + 7*(84.5-74.833)² + 12*(64.5-74.833)²
               + 4*(64.5-74.833)² + 4*(54.5-74.833)² =  3896.7

Calculate the standard deviation
s = √(3896.7/29) = 11.592

Answer:
mean = 74.83
std. deviation = 11.59
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Approximate area under the curve f(x) =-x^2+2x+4 from x=0 to x=3 by using summation notation with six rectangles and use the the
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Answer:

Summation notation:

\frac{1}{2}\sum_{k=1}^6f((.5k))

or after using your function part:

\frac{1}{2}\sum_{k=1}^6(-(.5k)^2+2(.5k)+4)

After evaluating you get 11.125 square units.

Step-by-step explanation:

The width of each rectangle is the same so we want to take the distance from x=0 to x=3 and divide by 6 since we want 6 equal base lengths for our rectangles.

The distance between x=0 and x=3 is (3-0)=3.

We want to divide that length of 3 units by 6 which gives a length of a half per each base length.

We are doing right endpoint value so I'm going to stat at x=3. The first rectangle will be drawn to the height of f(3).

The next right endpoint is x=3-1/2=5/2=2.5, and the second rectangle will have a height of f(2.5).

The next will be at x=2.5-.5=2, and the third rectangle will have  a height of f(2).

The fourth rectangle will have a height of f(2-.5)=f(1.5).

The fifth one will have a height of f(1.5-.5)=f(1).

The last one because it is the sixth one will have a height of f(1-.5)=f(.5).

So to find the area of a rectangle you do base*time.

So we just need to evaluate:

\frac{1}{2}f(3)+\frac{1}{2}f(2.5)+\frac{1}{2}f(2)+\frac{1}{2}f(1.5)+\frac{1}{2}f(1)+\frac{1}{2}f(.5)

or by factoring out the 1/2 part:

\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

To find f(3) replace x in -x^2+2x+4 with 3:

-3^2+2(3)+4

-9+6+4

1

To find f(2.5) replace x in -x^2+2x+4 with 2.5:

-2.5^2+2(2.5)+4

-6.25+5+4

2.75

To find f(2) replace x in -x^2+2x+4 with 2:

-2^2+2(2)+4

-4+4+4

4

To find (1.5) replace x in -x^2+2x+4 with 1.5:

-1.5^2+2(1.5)+4

-2.25+3+4

4.75

To find f(1) replace x in -x^2+2x+4 with 1:

-1^2+2(1)+4

-1+2+4

5

To find f(.5) replace x in -x^2+2x+4 with .5:

-.5^2+2(.5)+4

-.25+1+4

4.75

Now let's add those heights.  After we obtain this sum we multiply by 1/2 and we have our approximate area:

\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

\frac{1}{2}(1+2.75+4+4.75+5+4.75)

\frac{1}{2}(22.25)

11.125

Okay now if you wanted the summation notation for:

\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

is it

\frac{1}{2}\sum_{k=1}^{6}(f(.5+.5(k-1)))

or after simplifying a bit:

\frac{1}{2}\sum_{k=1}^6 f((.5+.5k-.5))

\frac{1}{2}\sum_{k=1}^6f((.5k))

If you are wondering how I obtain the .5+.5(k-1):

I realize that 3,2.5,2,1.5,1,.5 is an arithmetic sequence with first term .5 if you the sequence from right to left (instead of left to right) and it is going up by .5 (reading from right to left.)

6 0
3 years ago
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