Mean, Median, Mode, Standard Deviation – Basic Statistics (Calculator)

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’s sample SD with n=5 => we divide 56 by (5-1)=4 => 56/4=14 => then sqrt(14) ~ 3.74 => about 3.7 as sample standard deviation.

Solution

The original sum for 10 values is 10 × 50=500. Removing that one value (call it x=70) leaves sum 500 - 70=430. Then the new mean for 9 values is 430/9= about 47.8 (but the question states 47, so let's see if the question is approximate or if x differs).
The question specifically says the new mean is 47 for the 9 values. Then the sum of the 9 is 9×47=423 => so x= 500-423=77. Actually that suggests the removed value is 77 if the new mean is exactly 47.
The question asks for the removed value’s deviation from original mean => the original mean=50 => if the removed value is 77 => that deviation is 77-50= +27.
Probably the correct removed value is 77, not 70, given the question’s statement about the new mean. The question might have a mismatch. But presumably the correct approach is: sum originally= 500 => new sum= 9×47=423 => removed value= 500-423=77 => deviation= 77-50= 27.

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 70,80,85 plus unknown x. Let’s consider possible sorted scenarios.
If x≥85 => sorted would be 70,80,85,x => the 2nd,3rd are 80,85 => median=82.5 => that matches. So x≥85 works for that scenario.
If 80≤ x≤85 => sorted might be 70, x, 80, 85 or 70,80,x,85 => we must ensure the 2nd,3rd are the correct. Actually if x≥80, it might move the second or third positions. Let's be systematic: - If x≥80 but <85 => sorted => 70,80,x,85 => 2nd=80, 3rd=x => median= (80+x)/2 => we want 82.5 => 80+x=165 => x=85 => but that contradicts x<85. So no solution there. - If x≥85 => we got median=82.5 from 80 and 85 => that works.
If x<80 => sorted => 70, x, 80, 85 => the 2nd,3rd => x,80 => median= (x+80)/2=82.5 => x+80=165 => x=85 => but that contradicts x<80 => no solution.
So x≥85 is the set of possible values for x.

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 data by a constant “a” multiplies the mean by that constant and also multiplies the standard deviation by the absolute value of that constant.
So new mean= 3×60=180, new standard deviation= 3×5=15.

Solution

For 6 data points, the median is the average of the 3rd and 4th in sorted order=10 => that means the average(3rd,4th)=10. Possibly 3rd=10,4th=10 or 3rd=9,4th=11 or other combos. But adding 25, we have 7 data points => median is the 4th value => 11.5 => so 4th value= 11.5 if sorted.
That suggests that the new data’s 4th position is >10 for sure, so likely at least 3 of the original numbers were ≥ 10. The details can vary, but we can deduce that among the original, the 4th was at least 10, possibly the 3rd also.
The problem is open ended; typically we'd deduce at least half are ≥10. Because the median is the boundary. With 6 data points, 3rd or 4th is around 10. Usually means at least 2 of them are ≥10 or 3. Possibly the question suggests at least 3 are ≥10 because with the new data 25, the median jumps to 11.5.

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

For 5 data points in ascending order, the 3rd is the median => which is 6. The set so far is 2, 5, 6, 7, x. In ascending order, 6 must remain 3rd. That means 6≥5 => that’s consistent. We need x≥7 for 7 to remain 4th and x to be 5th. Or x could slot between 6 and 7. Actually if x≥7 => then the sorted would be 2,5,6,7,x => median=6 => valid. If x<7 but x≥6 => x might push the order. If x=6 => sorted is 2,5,6,6,7 => still median=6 => works. If x<6 => then the 3rd element might shift => possibly that breaks the median. So x≥6 => that keeps 6 as the 3rd item. So x≥6. But x≥7 or x=6 => all satisfy. So integer x≥6 => all works, i.e. x=6,7,8,9,...

Solution

Original sum= 10+10+12+14+16+20+20=102 => 7 data points => mean=102/7= about 14.57 but the question states the mean is 14 => check: 102/7=14.57 => conflict. Possibly the question approximates or there's a mismatch. Let’s see if the question is approximate or we can correct. Possibly the mean is ~14.57, or maybe there's a missing detail. We'll proceed.
Suppose the sum is 102, plus the new data 14 => sum=116 => new count=8 => new mean=116/8=14.5.
The new dataset sorted= 10,10,12,14,14,16,20,20 => median is average of 4th,5th => (14+14)/2=14 => mode is 10 and 14 and 20 each appear twice => so we have three modes.

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) We do population formula. Deviation from mean => 2-4.4= -2.4 => squared=5.76 => next 2 => same => 5.76 => 3 => -1.4 => squared=1.96 => 5 => +0.6 => squared=0.36 => 10 => +5.6 => squared=31.36 => sum=5.76+5.76+1.96+0.36+31.36=45.2 => population variance= 45.2/5=9.04 => std dev= sqrt(9.04)=3.0 approx.

Solution

5 appears twice, 7 appears twice. If x=5 => 5 would appear 3 times => definitely mode=5. If x=7 => 7 also appears 3 times => tie => 5 wouldn't be sole mode. So x cannot be 7. If x=4 => 4 appear once => doesn't surpass 5 => 5 remains a mode but so does 7. Actually 7 appears 2 times, 5 appears 2 times => tie. We want 5 to be the single mode. So no other number can appear 2+ times. 7 is already 2 times => if x=7 => that makes 7 have 3 times => overshadow 5? Actually that would overshadow 5 or tie? Actually it would overshadow 5 if x=7 => 7 is 3 times? Wait we see 7 is currently 2 times => plus x=7 => 3 times => that is more than 2 times => 7 becomes the sole mode => not allowed. So x=5 => 5 becomes 3 times => 7 is 2 times => that’s good => 5 is mode. If x=any other integer that is not 5 or 7 => that new number is 1 time => 7 is 2 times => 5 is 2 times => we get tie. So 5 wouldn't be the single mode. We want 5 to remain the single mode => must be x=5 => 5 appears 3 times while 7 only 2 times => 4 once => so 5 is single mode.
So 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

Adding a large outlier only shifts the upper portion but typically Q1, Q2, Q3 might remain the same if the outlier doesn’t shift how many data points are above or below. Usually quartiles remain unaffected if the new data is above Q3, unless the sample size is small enough that the ranks shift. But generally for a large sample, Q1, Q2, Q3 remain the same or barely shift.

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 IQR= Q3-Q1= 12-5=7 => that remains the same if Q1, Q3 do not change. The median also remains 8 if the data set is large enough that the maximum's shift does not affect quartiles or median. So the IQR stays 7, median still 8.

Solution

Typically, if the data are the entire population, we use population standard deviation \(\sigma\). If it’s a sample from a larger population, we use sample standard deviation s. The SAT often doesn’t specify which but in advanced contexts, “sample” means dividing by (n-1).

Solution

If x<4 => then we have only two 4’s, which is presumably more frequent than any other. So 4 remains the mode. If x=4 => we have three 4’s => 4 definitely remains the mode (with an even stronger frequency).

Solution

Removing 10 lowers the sum, likely lowers the mean. The new set 1,2,5,8 => sum=16 => mean=4 => that’s smaller. The standard deviation also typically decreases because the outlier is removed and data is more clustered.

Solution

We have a=5, b+c=17 (since b+c=2×8.5=17). Also a+b+c+d=36 => 5+(b+c)+d=36 => 5+17+d=36 => d=14. Then b+c=17 => we can choose b≤c => many combos: (8,9), (7,10), etc. Must keep sorted => 5≤ b ≤ c ≤14. For example b=8, c=9 => that works => set= {5,8,9,14}.

Solution

The transformations are: first +10, then ×2 => overall each original x -> 2(x+10)= 2x+20.
The mean transforms the same way => new mean= 2( old mean +10)= 2(20+10)=2(30)=60.
The standard deviation is unaffected by +10 but multiplied by 2 => new std dev= 2×4=8.

Solution

Mean= (3+5+5+7)=20/4=5 => deviations => 3-5= -2 => squared=4; 5-5=0 => squared=0; 5-5=0 => squared=0; 7-5=2 => squared=4 => sum=8 => sample variance= 8/(4-1)= 8/3 => 2.6667 => s= sqrt(2.6667)= about1.63.

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

Original sum= (6+6+6+7+8)=33 => 5 data points => mean=6.6 => new sum=33+2=35 => 6 points => new mean=35/6= about 5.83 => new median => sorted => {2,6,6,6,7,8} => median is average of 3rd,4th => (6+6)/2=6 => so the median decreases from 6 to 6 (coincidentally 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 4th and 5th. That average is 10. Adding x=10 => 9 data points => the 5th is the median => likely remains 10. Typically the median won't shift if the new data is at or near the existing median.

Solution

For right-skewed: mode < median < mean => 50k < 55k < 60k.

Solution

Left-skewed (negatively skewed) typically has mean < median < mode. The tail is on the left side, pulling mean below the median.

Solution

For 6 numbers, median= average of 3rd and 4th => (4 + 6)/2=5 => that’s consistent with 4,6. So presumably x≥6 => for the order to remain 2,4,4,6,x,8 => x≥6 => but that might not change the median. The question states the median is 5 => that’s from the 3rd,4th => 4,6 => average=5 => consistent. So x≥6. Possibly x could be 7 or 10 or anything≥6.

Solution

The transformation is: x -> (x - 14)/2. Subtracting 14 from each shifts mean by -14 => new mean= (original mean -14)/2= (14-14)/2=0/2=0.
For standard deviation, dividing by 2 => new st.dev= (original st.dev)/2= 2.83/2= about 1.415.