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Crazy boy [7]
3 years ago
9

Each year, taxpayers are able to contribute money to various charities via their IRS tax forms. The following list contains the

amounts of money (in dollars) donated via IRS tax forms by Each year, taxpayers are able to contribute money taxpayers:
2 , 22 , 27 , 31 , 36 , 51 , 57 , 57 , 60 , 62 , 62 , 62 , 73 , 77 , 83 , 95 , 99 , 104 , 105 , 127 , 153 , 162 , 197

(a) For these data, which measures of central tendency take more than one value? Choose all that apply.

Mean

Median

Mode

None of these measures

(b) Suppose that the measurement 197 (the largest measurement in the data set) were replaced by 246. Which measures of central tendency would be affected by the change? Choose all that apply.

Mean

Median

Mode

None of these measures

(c) Suppose that, starting with the original data set, the largest measurement were removed. Which measures of central tendency would be changed from those of the original data set? Choose all that apply.

Mean

Median

Mode

None of these measures

(d) Which of the following best describes the distribution of the original data? Choose only one.

Negatively skewed

Positively skewed

Roughly symmetrical

Mathematics
2 answers:
user100 [1]3 years ago
6 0

Answer:

(a) None of these measures

(b) Mean

(c) Mean and Median

(d) Roughly Symmetrical

Step-by-step explanation:

(a)

Mean

Total number in the set = 23

Summation of the set = 2+22+27+31+36+51+57+57+60+62+62+62+73+77+83+95+99+104+105+127+153+162+197 = 1804

Mean = Sum of set / total no of set

1804/23 = 78.435

Median is the middle number in the set after it had been arranged from lowest to highest

2 , 22 , 27 , 31 , 36 , 51 , 57 , 57 , 60 , 62 , 62 , 62 , 73 , 77 , 83 , 95 , 99 , 104 , 105 , 127 , 153 , 162 , 197

The Median is 62

Mode the value that appear most

Mode is 62

None of them takes more than one value

(b) If 197 is replaced by 246, the set becomes

2 , 22 , 27 , 31 , 36 , 51 , 57 , 57 , 60 , 62 , 62 , 62 , 73 , 77 , 83 , 95 , 99 , 104 , 105 , 127 , 153 , 162 , 246

The mean becomes

Total number in the set = 23

Summation of the set = 2+22+27+31+36+51+57+57+60+62+62+62+73+77+83+95+99+104+105+127+153+162+246= 1853

Mean = Sum of set / total no of set

1853/23 = 80.565

The Median and Mode remains the same.

(c) When the largest measurements are removed, the number of values in the set reduces and this affects the Mean and the Median. The mode will still remain unchanges since it is a small number and appears the most.

monitta3 years ago
3 0

Answer:

A. median and mode

B. Mean

C. Mean

d. Positively skewed

Step-by-step explanation:

a. Mean = sum of all values / numbers of values = 1804 / 23 = 78.4. this eliminates the mean and leaves the median and mode, both of which are 62.  

Median = middlemost value = 62

Mode = most frequent value = 62

b. Only the Mean would be affected. It would move from 78.4 to 80.6. (calculated as 1853 / 23). The median and mode would not be affected as the 62 would remain the middlemost value and the most frequent.

 

c. Removing the largest value would result in the mean changing. The mean would now be 73 (calculated as 1607 / 22). Since the values are no longer 23, but now 22, the median would be the average of the 2 middlemost values, ((62+62) / 2) which would still result in 62 as the median.  

d. we use the Five number summary:  

Median / Interquartile Range = 62

First Quartile / Q1 = (51 +57) / 2 = 54

Third Quartile / Q3 =  (99 + 104)  / 2 = 101.5

Minimum = 2

Maximum = 197 ; therefore Range = (197 - 2) = 195.

Please see attached box and whisker plot. The distribution is positively skewed.

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A selective college would like to have an entering class of 950 students. Because not all students who are offered admission acc
pogonyaev

Answer:

a) The mean is 900 and the standard deviation is 15.

b) 100% probability that at least 800 students accept.

c) 0.05% probability that more than 950 will accept.

d) 94.84% probability that more than 950 will accept

Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

(a) What are the mean and the standard deviation of the number X of students who accept?

n = 1200, p = 0.75. So

E(X) = np = 1200*0.75 = 900

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1200*0.75*0.25} = 15

The mean is 900 and the standard deviation is 15.

(b) Use the Normal approximation to find the probability that at least 800 students accept.

Using continuity corrections, this is P(X \geq 800 - 0.5) = P(X \geq 799.5), which is 1 subtracted by the pvalue of Z when X = 799.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{799.5 - 900}{15}

Z = -6.7

Z = -6.7 has a pvalue of 0.

1 - 0 = 1

100% probability that at least 800 students accept.

(c) The college does not want more than 950 students. What is the probability that more than 950 will accept?

Using continuity corrections, this is P(X \geq 950 - 0.5) = P(X \geq 949.5), which is 1 subtracted by the pvalue of Z when X = 949.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{949.5 - 900}{15}

Z = 3.3

Z = 3.3 has a pvalue of 0.9995

1 - 0.9995 = 0.0005

0.05% probability that more than 950 will accept.

(d) If the college decides to increase the number of admission offers to 1300, what is the probability that more than 950 will accept?

Now n = 1300. So

E(X) = np = 1300*0.75 = 975

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1200*0.75*0.25} = 15.6

Same logic as c.

Z = \frac{X - \mu}{\sigma}

Z = \frac{949.5 - 975}{15.6}

Z = -1.63

Z = -1.63 has a pvalue of 0.0516

1 - 0.0516 = 0.9484

94.84% probability that more than 950 will accept

5 0
3 years ago
Answer the following questions:
Lelechka [254]

Answer:

a. RTP

b. NM

c. x = 7

3x - 1 = 20

3x = 21

x = 7

5 0
3 years ago
If WZYX is equal to PMLN describes two quadrilaterals, which other statement is also true? I NEED THIS ANSWER ASAP
kondor19780726 [428]

Answer:

C.WXYZ equal to PNLM

Step-by-step explanation:

Look at the given statement, WZYX is equal to PMLN.

WZYX is equal to PMLN.   W corresponds to P.

WZYX is equal to PMLN.   Z corresponds to M.

WZYX is equal to PMLN.   Y corresponds to L.

WZYX is equal to PMLN.   X corresponds to N

Now look in the choices. The letters must correspond like they do above.

W must correspond to P.

WXYZ equal to P...

X must correspond to N.

WXYZ equal to PN...

Y must correspond to L.

WXYZ equal to PNL...

Z must correspond to M.

WXYZ equal to PNLM

Answer: C.WXYZ equal to PNLM

8 0
3 years ago
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The weight of a mandarin orange is about 6 ounces. A basket that weighs 4 ounces will hold m mandarin oranges, the total weight
trasher [3.6K]

Answer:

4 + 6m = 70

Step-by-step explanation:

4 is the basket

6m is the total weight of mandarins

70 is total weight

7 0
3 years ago
Can anyone help me with this please
tigry1 [53]
It’s b here is proof of what i got
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