Solution

(a) The mean is the sum of all scores divided by the number of students. Sum = 70 + 75 + 80 + 85 + 90 = 400. There are 5 scores, so mean = 400 / 5 = 80.

(b) Because there are 5 scores listed in ascending order, the median is the middle one, which is the 3rd score. That's 80.

(c) The mode is the most frequent score. In this list, all scores appear exactly once, so there is no single mode. Some texts might say "no mode" or "all are equally common."

Solution

First, let's sort them: 7, 8, 8, 9, 10, 10.

(a) Mean = sum / number of scores = (7 + 8 + 8 + 9 + 10 + 10) / 6 = 52 / 6 = 8.67 (approx).

(b) Since there are an even number (6) of data points, the median is the average of the 3rd and 4th values in sorted order. 3rd = 8, 4th = 9, so median = (8 + 9)/2 = 8.5.

(c) The values 8 and 10 each appear twice, more than the others which appear once each, so the modes are 8 and 10 (two modes).

Solution

Original 8 data points have a total sum = 8 × 12 = 96. When we add one new data point of 20, the sum becomes 96 + 20 = 116. Now we have 9 total data points. The new mean = 116 / 9 = about 12.89.

Solution

Because there are 7 numbers, the median is the 4th number in sorted order. The 4th number is m, and it's given that the median is 5.5, so m=5.5.

Solution

Adding a constant to each data point increases the mean by that constant but does not affect the standard deviation. So the new mean is 87.4 + 5 = 92.4, and the new standard deviation remains about 4.9.

Solution

(a) Mean = (2 + 4 + 6 + 6 + 12) / 5 = 30 / 5 = 6.

(b) For a sample standard deviation, we compute:
1) Deviations from mean: (2-6)= -4, (4-6)= -2, (6-6)=0, (6-6)=0, (12-6)=6.
2) Squares: 16, 4, 0, 0, 36 ⇒ sum= 56.
3) Because it is a sample standard deviation with \(n=5\), divide 56 by \(5-1=4\). The sample variance is \(56/4=14\), so \(s=\sqrt{14}\approx 3.74\). Rounded to the nearest tenth, the sample standard deviation is 3.7.

Solution

The original sum for 10 values is \(10 \times 50=500\). After one value is removed, the remaining 9 values have mean 47, so their sum is \(9 \times 47=423\).
The removed value is \(500-423=77\). Its deviation from the original mean is \(77-50=27\). The removed value was 27 units above the original mean.

Solution

The sum of the 5 values= 10+10+10+15+20=65 ⇒ mean= 65/5=13.
The median is the 3rd value in sorted list ⇒ that's 10.
The mode is 10, it appears 3 times.
The range= max-min= 20-10=10.

Solution

5 consecutive integers have a symmetrical distribution, so the mean is also the middle number. That middle number is the median. So the median is also 14. Another approach: The sum is 5×14=70 ⇒ let the 5 consecutive integers be n-2, n-1, n, n+1, n+2 ⇒ their sum= 5n ⇒ so 5n=70 ⇒ n=14 ⇒ so the median is 14.

Solution

For 4 data points in sorted order, the median is the average of the 2nd and 3rd points. The existing known scores are 70, 80, and 85, plus the unknown score \(x\).
If \(x \ge 85\), the sorted order is 70, 80, 85, \(x\). The middle two values are 80 and 85, so the median is \((80+85)/2=82.5\). This works.
If \(80 \le x < 85\), the sorted order is 70, 80, \(x\), 85. The median would be \((80+x)/2\). Setting this equal to 82.5 gives \(x=85\), which is outside this interval.
If \(x < 80\), the middle two values are \(x\) and 80. Setting \((x+80)/2=82.5\) gives \(x=85\), which is impossible in this case.
Therefore, the possible values are \(x \ge 85\).

Solution

The original mean is (2+5+7+7+9)/5=30/5=6. The median is the 3rd value (in sorted list) ⇒ 7. If we replicate each point, the new set is 10 data points but in the same sorted pattern (just doubled). The mean remains the same (6) because the sum doubles and the number of data points also doubles. The median, as we double them, the middle two values (for 10 points) remain around the central repeated 7. The median remains 7. So neither changes.

Solution

Multiplying every data value by a constant \(a\) multiplies the mean by \(a\) and multiplies the standard deviation by \(|a|\).
So the new mean is \(3 \times 60=180\), and the new standard deviation is \(3 \times 5=15\).

Solution

Write the original sorted values as \(a_1 After 25 is added, there are 7 values and the median is the 4th value. Because the new median is 11.5, the 4th value of the new sorted list is 11.5. In this setup, that value is \(a_4\), so \(a_4=11.5\). Then \(a_5\) and \(a_6\) are also greater than 11.5.
Therefore, exactly 3 of the original numbers are at least 10: \(a_4\), \(a_5\), and \(a_6\).

Solution

Original set mean is (3+5+5+7+10)/5=30/5=6 ⇒ some spread. Adding 3 adds a data point below the mean, which likely increases the spread overall because we have an additional low value. Generally, adding a data point far from the mean or on one side tends to increase standard deviation. So \(\sigma\) would increase.

Solution

Adding a constant does not change the spread. So the new standard deviation remains 2.

Solution

Remaining scores: 80, 81, 82, 83 ⇒ sum= 326 ⇒ average= 326/4=81.5.

Solution

Sum= 30+55+60+60+65=270 ⇒ mean=270/5=54 lb.
The mode is 60 because it appears twice, more than others.
If a 6th dog is also 60 ⇒ that reaffirms 60 is the mode, with now 3 occurrences.

Solution

All data points remain 4 ⇒ no variation ⇒ standard deviation is 0.

Solution

With 5 data points, the median is the 3rd value after sorting. The known values 2, 5, 6, and 7 already place 6 in the middle unless \(x\) is less than 6. If \(x<6\), then \(x\) would come before 6 and could move 6 out of the 3rd position. If \(x \ge 6\), the sorted set keeps 6 as the middle value.
Therefore, any integer \(x \ge 6\) works: \(6,7,8,9,\ldots\).

Solution

The original sum is \(10+10+12+14+16+20+20=102\). After adding 14, the new sum is \(116\) and the new count is 8, so the new mean is \(116/8=14.5\).
The new sorted dataset is 10, 10, 12, 14, 14, 16, 20, 20. The median is the average of the 4th and 5th values: \((14+14)/2=14\). The modes are 10, 14, and 20 because each appears twice.

Solution

Relationship: population std dev uses divisor n, sample std dev uses divisor n-1. So \(\sigma= \sqrt{\frac{n-1}{n}} \cdot s\). Here n=6 ⇒ \(\sigma= \sqrt{\frac{5}{6}} s= s\sqrt{\frac{5}{6}}.\)

Solution

Original sum=4×10=40. Adding 15,5 ⇒ new sum=40+15+5=60 ⇒ total 6 points ⇒ new mean=60/6=10 ⇒ so the mean remains 10, because 15+5=20 ⇒ that's exactly 2 times 10 ⇒ no net effect on average.

Solution

Original sum=5×12=60. The range= largest-min. If we add 6 to the largest, the new sum=60+6=66 ⇒ new mean=66/5=13.2 ⇒ range increases by 6 from the original 10 ⇒ new range=16.

Solution

For 4 values, median= average of 2nd,3rd ⇒ (7 + x)/2=8 ⇒ 7+x=16 ⇒ x=9. But also the list must remain sorted as 6,7,x,14 ⇒ so x≥7 ⇒ x≥7 ⇒ but we also have (7+x)/2=8 ⇒ x=9. We check if 9 <14 ⇒ yes ⇒ so x=9 is valid.

Solution

Repeating each data point does not change the distribution's spread. The standard deviation remains the same. Doubling the sample size in that way does not affect the distances among points; hence \(\sigma\) stays the same.

Solution

(a) Mean= (2+2+3+5+10)=22 /5=4.4.
(b) Sorted ⇒ 2,2,3,5,10 ⇒ 3rd is 3 ⇒ median=3.
(c) Use the population formula. The squared deviations from the mean 4.4 are \(5.76, 5.76, 1.96, 0.36,\) and \(31.36\). Their sum is 45.2. The population variance is \(45.2/5=9.04\), so the standard deviation is \(\sqrt{9.04}\approx 3.0\).

Solution

In the set {4, 5, 5, 7, 7, \(x\)}, both 5 and 7 already appear twice. For 5 to be the single mode, it must appear more often than 7.
If \(x=5\), then 5 appears 3 times while 7 appears 2 times. If \(x=7\), then 7 becomes the mode. If \(x\) is any other integer, 5 and 7 remain tied. Therefore, \(x=5\) is the only possibility.

Solution

Original sum= 7×50,000=350,000 ⇒ add one at 70,000 ⇒ new sum=420,000 ⇒ new average=420,000 ÷8=52,500.

Solution

A very large outlier affects the maximum, range, and mean most directly. Quartiles are based on position, so \(Q_1\), the median \(Q_2\), and \(Q_3\) often stay the same or change only slightly in a large dataset. In a very small dataset, the exact quartiles can shift because the number of data points changes.

Solution

Sum=1+2+4+9+16+25+25=82 ⇒ number of data points=7 ⇒ mean=82/7 ~11.71.
Sorted ⇒ 1,2,4,9,16,25,25 ⇒ the median is the 4th ⇒ 9.

Solution

The replacement changes only the maximum from 20 to 25. The quartiles and median listed in the 5-number summary remain \(Q_1=5\), median \(=8\), and \(Q_3=12\). Therefore, the IQR remains \(12-5=7\), and the median remains 8.

Solution

For population standard deviation, divide by \(n\): the variance is \(90/10=9\), so \(\sigma=3\). For sample standard deviation, divide by \(n-1\): the variance is \(90/(10-1)=10\), so \(s=\sqrt{10}\). Use the population formula when the data represent the whole group being studied; use the sample formula when the data are a sample from a larger group.

Solution

If \(x<4\), the value 4 still appears twice while every other value appears once, so 4 remains the mode. If \(x=4\), then 4 appears three times, so 4 remains the mode even more clearly.

Solution

Removing 10 lowers the sum and reduces the mean. The new set is 1, 2, 5, 8, with sum 16 and mean 4. The standard deviation decreases because the most extreme high value has been removed and the remaining values are more clustered.

Solution

Since the median is 8.5, \((b+c)/2=8.5\), so \(b+c=17\). The total sum is \(4 \times 9=36\). With \(a=5\), we get \(5+b+c+d=36\), so \(5+17+d=36\), which gives \(d=14\).
Therefore \(b\) and \(c\) can be any sorted pair with \(b+c=17\) and \(5 \le b \le c \le 14\). Examples include \(b=8, c=9\) or \(b=7, c=10\). One possible set is {5, 8, 9, 14}.

Solution

The transformation is: first add 10, then multiply by 2. Overall, each original value \(x\) becomes \(2(x+10)=2x+20\).
The mean transforms the same way, so the new mean is \(2(20+10)=60\).
The standard deviation is unaffected by adding 10, but it is multiplied by 2 when all values are doubled. The new standard deviation is \(2 \times 4=8\).

Solution

The deviations from the mean are \(-2, 0, 0,\) and \(2\). The squared deviations are \(4, 0, 0,\) and \(4\), so their sum is 8. For a sample with \(n=4\), divide by \(n-1=3\): \(s^2=8/3\approx 2.6667\). Therefore, \(s=\sqrt{2.6667}\approx 1.63\).

Solution

Adding a number below the old mean drags the average down ⇒ the new mean is lower than 6.

Solution

Range= max-min=15-2=13. The set has 8 data points ⇒ Q1 is average of 2nd and 3rd ⇒ (3+5)/2=4 ⇒ Q3 is average of 6th and 7th ⇒ (9+10)/2= 9.5 ⇒ IQR= Q3-Q1= 9.5-4=5.5.

Solution

Original median is the 4th number ⇒ 9. Removing 15 ⇒ new set {1,5,7,9,9,11} ⇒ 6 points ⇒ median is average of 3rd,4th ⇒ (7+9)/2=8 ⇒ the median decreases from 9 to 8.

Solution

Scaling by 1/5 ⇒ new mean= (1/5)*100=20 ⇒ new std dev= (1/5)*10=2.

Solution

If mode < median < mean, typically the distribution is skewed right (positively skewed). The tail extends to the right, pulling the mean above the median.

Solution

The original sum is \(6+6+6+7+8=33\), so the original mean is \(33/5=6.6\). After adding 2, the new sum is 35 and the new mean is \(35/6 \approx 5.83\).
The new sorted set is {2, 6, 6, 6, 7, 8}. With 6 values, the median is the average of the 3rd and 4th values: \((6+6)/2=6\). The mean decreases, but the median remains 6.

Solution

Typically, Q1 and Q3 remain the same or shift very little if the data set is large. A single outlier in the upper range typically does not drastically shift Q1 or Q3, especially if the sample is large. So Q1=9, Q3=15 likely remain or barely change.

Solution

Removing a low outlier typically decreases the spread, so the sample standard deviation likely decreases.

Solution

With 8 data points, the median is the average of the 4th and 5th values. If that average is 10, then the 4th value is at most 10 and the 5th value is at least 10. Adding a new value of 10 places 10 at the middle position of the 9-value set.
Therefore, the new median is 10.

Solution

For a right-skewed distribution, \(\text{mode} < \text{median} < \text{mean}\). In this case, \(\$50k < \$55k < \$60k\).

Solution

Left-skewed, or negatively skewed, data typically follow \(\text{mean} < \text{median} < \text{mode}\). The tail is on the left side, pulling the mean below the median.

Solution

For 6 numbers, the median is the average of the 3rd and 4th values. Here those values are 4 and 6, so the median is \((4+6)/2=5\).
Since the list is sorted in ascending order as 2, 4, 4, 6, \(x\), 8, the value of \(x\) must satisfy \(6 \le x \le 8\). If \(x\) is an integer, then \(x\) can be 6, 7, or 8.

Solution

The transformation is \(x \mapsto (x-14)/2\). The new mean is \((14-14)/2=0\).
Standard deviation is not affected by subtracting 14, but it is divided by 2 when all deviations are divided by 2. The new standard deviation is \(2.83/2 \approx 1.415\).