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3 years ago

5

You give a test to 100 students and determine the standard deviation of the scores. After grading the test, you realize that the

10 students with the highest scores did exceptionally well. You decide to award these 10 students a bonus of 5 more points each. The standard deviation of the new score distribution will be _______________________(less than, greater than or the same as) that of the original score distribution

1 answer:

Rasek [7]3 years ago

5 0

Answer:

greater than

Step-by-step explanation:

The standard deviation of a distribution σ = $\sqrt{sum(x - mean)^2/ n}$

where x is the score of each student

Adding 5 to 10 student scores would be = x+5 - mean

increasing the scores of the 10 students by 5, which in turn increases the standard deviation of the distribution.

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Factor the expression using the GCF. 6x + 42

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Answer: 6x + 42 +=78

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Jen deposited $750 in her savings account. The account earns 2% simple interest per year. She has the money in her account for 2

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2 years ago

Need help with homework

motikmotik

We have two points describing the diameter of a circumference, these are:

$\begin{gathered} A=(-12,-4) \\ B=(-4,-10) \end{gathered}$

Recall that the equation for the standard form of a circle is:

$(x-h)^2+(y-k)^2=r^2$

Where (h,k) is the coordinate of the center of the circle, to find this coordinate, we find the midpoint of the diameter, that is, the midpoint between points A and B.

For this we use the following equation:

$M=(\frac{x_1+x_2_{}_{}}{2},\frac{y_1+y_2}{2})$

Now, we replace and solve:

$\begin{gathered} M=(\frac{-12+(-4)}{2},\frac{-4+(-10)}{2} \\ M=(\frac{-12-4}{2},\frac{-4-10}{2}) \\ M=(\frac{-16}{2},\frac{-14}{2}) \\ M=(-8,-7) \end{gathered}$

The center of the circle is (-8,-7), so:

$\begin{gathered} h=-8 \\ k=-7 \end{gathered}$

On the other hand, we must find the radius of the circle, remember that the radius of a circle goes from the center of the circumference to a point on its arc, for this we use the following equation:

$r^2=\Delta x^2+\Delta y^2$

In this case, we will solve the delta with the center coordinate and the B coordinate.

$\begin{gathered} r^2=((-4)-(-8))^2+((-10)-(-7)) \\ r^2=(-4+8)^2+(-10+7)^2 \\ r^2=4^2+(-3)^2 \\ r^2=16+9 \\ r^2=25 \\ r=5 \end{gathered}$

Therefore, the equation for the standard form of a circle is:

$\begin{gathered} (x-(-8))^2+(y-(-7))^2=25 \\ (x+8)^2+(y+7)^2=25 \end{gathered}$

In conclusion, the equation is the following:

$(x+8)^2+(y+7)^2=25$

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10 months ago

It takes Kevin 15 minutes to drive 52 miles.

Kaylis [27]

Answer:

C

Step-by-step explanation:

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2 years ago

A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mat

Helen [10]

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

Step-by-step explanation:

1) Data given and notation

$s^2 =16$ represent the sample variance

s=4 represent the sample standard deviation

$\bar x$ represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

$\chi^2_{\alpha/2}=36.191$

$\chi^2_{1- \alpha/2}=7.633$

And replacing into the formula for the interval we got:

$\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}$

$8.400 \leq \sigma^2 \leq 39.827$

So the 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

7 0

3 years ago