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2 years ago

Scores on a final exam taken by 1200 students have a bell shaped distribution with mean=72 and standard deviation=9

Mathematics

1 answer:

SVETLANKA909090 [29]2 years ago

8 0

Answer:

a. 72

b. 816

c. 570

d. 30

Step-by-step explanation:

Given the graph is a bell - shaped curve. So, we understand that this is a normal distribution and that the bell - shaped curve is a symmetric curve.

Please refer the figure for a better understanding.

a. In a normal distribution, Mean = Median = Mode

Therefore, Median = Mean = 72

b. We have to know that 68% of the values are within the first standard deviation of the mean.

i.e., 68% values are between Mean $\pm$ Standard Deviation (SD).

Scores between 63 and 81 :

Note that 72 - 9 = 63 and

72 + 9 = 81

This implies scores between 63 and 81 constitute 68% of the values, 34% each, since the curve is symmetric.

Now, Scores between 63 and 81 = $\frac{68}{100} \times 1200$

= 68 X 12 = 816.

That means 816 students have scored between 63 and 81.

c. We have to know that 95% of the values lie between second Standard Deviation of the mean.

i.e., 95% values are between Mean $\pm$ 2(SD).

Note that 90 = 72 + 2(9) = 72 + 18

Also, 54 = 63 - 18.

Scores between 54 and 90 totally constitute 95% of the values. So, Scores between 72 and 90 should amount to $\frac{95}{2} \%$ of the values.

Therefore, Scores between 72 and 90 = $\frac{95}{2(100)} \times 1200 = \frac{95}{200} \times 1200$

$\implies 95 \times 12$ = 570.

That is a total of 570 students scored between 72 and 90.

d. We have to know that 5 % of the values lie on the thirst standard Deviation of the mean.

In this case, 5 % of the values lie between below 54 and above 90.

Since, we are asked to find scores below 54. It should be 2.5% of the values.

So, Scores below 54 = $\frac{2.5}{100} \times 1200$

= 2.5 X 12 = 30.

That is, 30 students have scored below 54.

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We need to find out the effects on the y-intercept when shifting the function into:

We know that the graph intersects the y-axis when , therefore:

So:

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An example is shown in Figure 1. The graph in blue is the function:

and the function in red is:

So you can see that:

2.2. Effects on the regions where the graph is increasing and decreasing

The effects are the same just as in the previous case. So the new function increases and decreases in the same intervals of

In Figure 1 you can see that both functions increase at:

and decrease at:

2.3 The end behavior when the following changes are made.

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FOR EVEN FUNCTIONS:

3. When becomes

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We need to find out the effects on the y-intercept when shifting the function into:

As we know, the graph intersects the y-axis when , therefore:

And:

So the new y-intercept is the negative of the previous intercept shifted one unit upward.

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In the intervals when the function increases, the function decreases. On the other hand, in the intervals when the function decreases, the function increases.

3.3 The end behavior when the following changes are made.

Each point of the function has the same x-coordinate just as the function and the y-coordinate is the negative of the previous coordinate shifted one unit upward, that is:

FOR ODD FUNCTIONS:

4. When becomes

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4.2. Effects on the regions where the graph is increasing and decreasing

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4.3 The end behavior when the following changes are made.

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