How to find median - PSAT Math
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The average of 5, 10, 12, 15, and x is 11. What is the median?
The average of 5, 10, 12, 15, and x is 11. What is the median?
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You must first find what x is in order to find the median.
To find x, set up the following equation:
To solve the equation, first multiply both sides by 5:
5 + 10 + 12 + 15 + x = 55
Then, add up 5 + 10 + 12 + 15 to get 42:
42 + x = 55
x = 13
Now that you know what x is, you are ready to find the median. To find the median, order the numbers from lowest to highest. The median is the number in the middle.
5, 10, 12, 13, 15
You must first find what x is in order to find the median.
To find x, set up the following equation:
To solve the equation, first multiply both sides by 5:
5 + 10 + 12 + 15 + x = 55
Then, add up 5 + 10 + 12 + 15 to get 42:
42 + x = 55
x = 13
Now that you know what x is, you are ready to find the median. To find the median, order the numbers from lowest to highest. The median is the number in the middle.
5, 10, 12, 13, 15
The chart above shows the scores on two tests for five students in a science class. If John's score was equal to the mode for Test I and equal to the median for Test II, then which of the five students represents John's scores?
The chart above shows the scores on two tests for five students in a science class. If John's score was equal to the mode for Test I and equal to the median for Test II, then which of the five students represents John's scores?
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The mode of a set of numbers is the number that appears most frequently in that set. The median of a set of numbers is the number that appears in the middle when they are arranged in numerical order. For Test I, the mode is 85. When arranged in numerical order, the scores for Test II are:
70, 70, 80, 85, 90.
That makes the median for Test II 80. The only student in the chart with an 85 on Test I and an 80 on Test II is student B.
The mode of a set of numbers is the number that appears most frequently in that set. The median of a set of numbers is the number that appears in the middle when they are arranged in numerical order. For Test I, the mode is 85. When arranged in numerical order, the scores for Test II are:
70, 70, 80, 85, 90.
That makes the median for Test II 80. The only student in the chart with an 85 on Test I and an 80 on Test II is student B.
The average (arithmetic mean) of 1, 7, 9, and n is –1. What is the median of this set?
The average (arithmetic mean) of 1, 7, 9, and n is –1. What is the median of this set?
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If the mean is –1, then 1 + 7 + 9 + n = –4. Solving for n gives us n = –21. In numeric order, we now have –21, 1, 7, 9. As there is an even number of numbers, we average the middle two. (1 + 7) / 2 = 4.
If the mean is –1, then 1 + 7 + 9 + n = –4. Solving for n gives us n = –21. In numeric order, we now have –21, 1, 7, 9. As there is an even number of numbers, we average the middle two. (1 + 7) / 2 = 4.
The monthly averaged high-and low-temperatures in Michigan are shown in the given table. What are the median monthly averaged high- and low-temperatures respectively in Michigan?
The monthly averaged high-and low-temperatures in Michigan are shown in the given table. What are the median monthly averaged high- and low-temperatures respectively in Michigan?
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Median is the middle number when the data are ordered from least to greatest. In this case, there are 12 numbers, so, the median is the average of the middle two numbers, which are the 6th and 7th numbers in an ascending order. For the averaged high-temperature, the median is (56+60)/2=58, and for the averaged low-temperature, the median is (35+39)/2=37
Median is the middle number when the data are ordered from least to greatest. In this case, there are 12 numbers, so, the median is the average of the middle two numbers, which are the 6th and 7th numbers in an ascending order. For the averaged high-temperature, the median is (56+60)/2=58, and for the averaged low-temperature, the median is (35+39)/2=37
Find the median of the following sequence:
3, 4, 5, 6, 7, 7, 10
Find the median of the following sequence:
3, 4, 5, 6, 7, 7, 10
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The median is the middle number in a sequence when the numbers are put in order. Since this sequence is already in order from least to greatest, we just need to find the middle number. There are 7 terms so the middle term is (7+1)/2 or the 4th term. This is 6.
The median is the middle number in a sequence when the numbers are put in order. Since this sequence is already in order from least to greatest, we just need to find the middle number. There are 7 terms so the middle term is (7+1)/2 or the 4th term. This is 6.
Find the median of the data set:
25, 37, 13, 58, 52, 83, 21, 51
Find the median of the data set:
25, 37, 13, 58, 52, 83, 21, 51
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42.5 is the mean of the data. 13 is the minimum. 83 is the maximum. 70 is the range.
To find the median, list all numbers in order:
13, 21, 25, 37, 51, 52, 58, 83
and find the middle value. In cases like this where there are two middle numbers (37 and 51), find the mean of these two numbers.
(37+51)/2 = 88/2 = 44
42.5 is the mean of the data. 13 is the minimum. 83 is the maximum. 70 is the range.
To find the median, list all numbers in order:
13, 21, 25, 37, 51, 52, 58, 83
and find the middle value. In cases like this where there are two middle numbers (37 and 51), find the mean of these two numbers.
(37+51)/2 = 88/2 = 44
The Brenner cousins' heights, in inches, are as follows:
Jeremy: 64
Vanessa: 69
Tracie: 60
Samuel: 70
Raymond: 74
Justin: 72
Patty: 55
Lauren: 52
Keith: 58
What is the median height of the cousins?
The Brenner cousins' heights, in inches, are as follows:
Jeremy: 64
Vanessa: 69
Tracie: 60
Samuel: 70
Raymond: 74
Justin: 72
Patty: 55
Lauren: 52
Keith: 58
What is the median height of the cousins?
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To find the median, one must arrange all the heights from the lowest to the highest value and then pick the middle value.
All values: 64 69 60 70 74 72 55 52 58
In order from lowest: 52 55 58 60 64 69 70 72 74
Median: 64
To find the median, one must arrange all the heights from the lowest to the highest value and then pick the middle value.
All values: 64 69 60 70 74 72 55 52 58
In order from lowest: 52 55 58 60 64 69 70 72 74
Median: 64
Find the median in this set of numbers
2, 100, 52, 97, 1, 7, 22
Find the median in this set of numbers
2, 100, 52, 97, 1, 7, 22
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in order to find the median, arrange the numbers in ascending order, and find the number in the middle of the list
in order to find the median, arrange the numbers in ascending order, and find the number in the middle of the list
Find the median from the given set of numbers
1, 4, 8, 17, 8, 8, 15, 21, 32, 17
Find the median from the given set of numbers
1, 4, 8, 17, 8, 8, 15, 21, 32, 17
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In order to find the median, arrange the numbers from smallest to biggest and find the number in the middle of the set.
In this case, both the 8 and 15 are in the middle of the set; so take the average of those 2 numbers (add both and divide by 2)
In order to find the median, arrange the numbers from smallest to biggest and find the number in the middle of the set.
In this case, both the 8 and 15 are in the middle of the set; so take the average of those 2 numbers (add both and divide by 2)
The number n is to be added to the list {3, 4, 5, 6, 10, 12}. If n is an integer, which of the following could be the median of the new list of seven numbers?
I) 5
II) 5.5
III) 6
The number n is to be added to the list {3, 4, 5, 6, 10, 12}. If n is an integer, which of the following could be the median of the new list of seven numbers?
I) 5
II) 5.5
III) 6
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Before n is added to the list, the median is 5.5 (the average of 5 and 6). When n is added to the list, the number of elements becomes odd, so the median will be a value directly from the list, not the average of two values. All of the values in the old list are integers and n is an integer, so the new median must be an integer; therefore, 5.5 cannot be the median of the new list.
Considering some possible values of n, we see that in cases where n is less than or equal to 5, the fourth element in the new list would be 5, making the new median 5. In cases where n is greater than or equal to 6, the fourth element in the new list would be 6, making the new median 6. The possible values for the median of the new list are therefore 5 and 6, but not 5.5.
Before n is added to the list, the median is 5.5 (the average of 5 and 6). When n is added to the list, the number of elements becomes odd, so the median will be a value directly from the list, not the average of two values. All of the values in the old list are integers and n is an integer, so the new median must be an integer; therefore, 5.5 cannot be the median of the new list.
Considering some possible values of n, we see that in cases where n is less than or equal to 5, the fourth element in the new list would be 5, making the new median 5. In cases where n is greater than or equal to 6, the fourth element in the new list would be 6, making the new median 6. The possible values for the median of the new list are therefore 5 and 6, but not 5.5.
M is a set consisting of a finite number of consecutive integers. If the average (arithmetic mean) of the numbers in set M is equal to one of the numbers in set M, which of the following must be true?
I. The number of numbers in set M is odd.
II. The average (arithmetic mean) of the numbers in set M equals the median.
III. Set M has a unique mode.
M is a set consisting of a finite number of consecutive integers. If the average (arithmetic mean) of the numbers in set M is equal to one of the numbers in set M, which of the following must be true?
I. The number of numbers in set M is odd.
II. The average (arithmetic mean) of the numbers in set M equals the median.
III. Set M has a unique mode.
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Statement I must be true because if M had an even number of consecutive integers, then its average (and median) would be the average of two consecutive integers, which is a decimal value rather than an integer, and therefore cannot be in set M, which contains only integers.
To check Statement II, consider some simple possible sets for M such as {0,1,2}. We see that in a set with an odd number of items, like M, the median is always the middle element. We also see that in a set with an odd number of consecutive integers (such as {0,1,2}), the average of the set will always be the exact middle element, too. Therefore, the average and the median must be equal, and Statement II must be true.
Finally, we can find a counterexample to statement III to show that it does not have to be true. If set M is {0,1,2}, we see that there is no unique mode (in fact, the only time set M could have a unique mode is when it only has one element!)
Statement I must be true because if M had an even number of consecutive integers, then its average (and median) would be the average of two consecutive integers, which is a decimal value rather than an integer, and therefore cannot be in set M, which contains only integers.
To check Statement II, consider some simple possible sets for M such as {0,1,2}. We see that in a set with an odd number of items, like M, the median is always the middle element. We also see that in a set with an odd number of consecutive integers (such as {0,1,2}), the average of the set will always be the exact middle element, too. Therefore, the average and the median must be equal, and Statement II must be true.
Finally, we can find a counterexample to statement III to show that it does not have to be true. If set M is {0,1,2}, we see that there is no unique mode (in fact, the only time set M could have a unique mode is when it only has one element!)
If x > 0, what is the median of the set of numbers?
{x + 2, x – 4, x + 6, x – 8}
If x > 0, what is the median of the set of numbers?
{x + 2, x – 4, x + 6, x – 8}
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To find the median, you need to find the middle number of the ordered list. Since there are four numbers in the list, you need to average the two middle numbers.
Ordered list = {x – 8, x – 4, x + 2, x + 6}
((x – 4) + (x + 2))/2 = (2x – 2)/2 = x – 1
To find the median, you need to find the middle number of the ordered list. Since there are four numbers in the list, you need to average the two middle numbers.
Ordered list = {x – 8, x – 4, x + 2, x + 6}
((x – 4) + (x + 2))/2 = (2x – 2)/2 = x – 1
A car travels at 60 miles per hour for 3 hours, 55 miles per hour for 2 hours, and 40 miles per hour for 3 hours. What (to the closest hundreth) is its average speed over the whole course of this trip?
A car travels at 60 miles per hour for 3 hours, 55 miles per hour for 2 hours, and 40 miles per hour for 3 hours. What (to the closest hundreth) is its average speed over the whole course of this trip?
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The easiest way to solve this is to find the total number of miles traveled by the car and divide that by the total time travelled.
Recall that D = rt; therefore, for each of these three periods, we can calculate the distance and sum those products:
Dtotal = 60 * 3 + 55 * 2 + 40 * 3 = 180 + 110 + 120 = 410
The total amount of time travelled is: 3 + 2 + 3 = 8
Therefore, the average rate is 410 / 8 = 51.25 miles per hour.
The easiest way to solve this is to find the total number of miles traveled by the car and divide that by the total time travelled.
Recall that D = rt; therefore, for each of these three periods, we can calculate the distance and sum those products:
Dtotal = 60 * 3 + 55 * 2 + 40 * 3 = 180 + 110 + 120 = 410
The total amount of time travelled is: 3 + 2 + 3 = 8
Therefore, the average rate is 410 / 8 = 51.25 miles per hour.
Set S consists of a set of six positive consecutive odd integers. Set T consists of the squares of each of the six elements in S. If the range of T is 560, then what is the median of T?
Set S consists of a set of six positive consecutive odd integers. Set T consists of the squares of each of the six elements in S. If the range of T is 560, then what is the median of T?
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Let's let a and b stand for the smallest and largest values in S, respectively. Because T is equal to the square of the elements in S, then the smallest and largest values in T will be equal to _a_2 and _b_2 , respectively.
We are told that the range of T is equal to 560. The range of a set of numbers is the difference between the largest and smallest, which in this case we can represent as _b_2 – _a_2. We can now write the following equation:
_b_2 – _a_2 = 560
Noice that _b_2 – _a_2 can be factored as (b – a)(b + a), which is the formula for the difference of squares.
(b – a)(b + a) = 560
Notice that the value b – a represents the range of S, because it is the difference between the largest and smallest values of S.
We are told that S consists of six consecutive odd integers. We can use this information to find the range of S. Because they are consecutive odd integers, each member in S will be two larger than the one before it. Thus, we could represent S as follows:
a, a + 2, a + 4, a + 6, a + 8, a + 10
The range of S is equal (a + 10) – a = 10. Thus, b – a must equal 10.
Let's go back to the equation (b – a)(b + a) = 560. We can replace b – a with 10 and solve for b + a.
(10)(b + a) = 560
Divide both sides by 10.
b + a = 56
So, we know that b – a = 10, and b + a = 56. We can solve this system of equations to find the values of a and b, which will give us the smallest and largest values in S.
Let's solve the first equation in terms of b.
b – a = 10
Add a to both sides.
b = 10 + a
Now we can substitute 10 + a in for b in the second equation.
(10 + a) + a = 56
10 + 2_a_ = 56
Subtract 10 from both sides.
2_a_ = 46
Divide both sides by 2.
a = 23
b = 10 + a = 10 + 23 = 33
The smallest value of S is 23, and the largest value is 33. Thus, S consists of the following set:
23, 25, 27, 29, 31, 33
We are told that T is equal to the squares of the elements of S. Thus, T consists of the following numbers:
529, 625, 729, 841, 961, 1089
We are asked ultimately to find the median of T. The median of any set is the middle number. In this case, T has six elements, so there are actually two numbers in the middle: 729 and 841. To find the median, we will take the average of the two middle numbers. The average of two numbers is equal to their sum divided by two.
median of T = (729 + 841)/2 = 1570/2 = 785.
The answer is 785.
Let's let a and b stand for the smallest and largest values in S, respectively. Because T is equal to the square of the elements in S, then the smallest and largest values in T will be equal to _a_2 and _b_2 , respectively.
We are told that the range of T is equal to 560. The range of a set of numbers is the difference between the largest and smallest, which in this case we can represent as _b_2 – _a_2. We can now write the following equation:
_b_2 – _a_2 = 560
Noice that _b_2 – _a_2 can be factored as (b – a)(b + a), which is the formula for the difference of squares.
(b – a)(b + a) = 560
Notice that the value b – a represents the range of S, because it is the difference between the largest and smallest values of S.
We are told that S consists of six consecutive odd integers. We can use this information to find the range of S. Because they are consecutive odd integers, each member in S will be two larger than the one before it. Thus, we could represent S as follows:
a, a + 2, a + 4, a + 6, a + 8, a + 10
The range of S is equal (a + 10) – a = 10. Thus, b – a must equal 10.
Let's go back to the equation (b – a)(b + a) = 560. We can replace b – a with 10 and solve for b + a.
(10)(b + a) = 560
Divide both sides by 10.
b + a = 56
So, we know that b – a = 10, and b + a = 56. We can solve this system of equations to find the values of a and b, which will give us the smallest and largest values in S.
Let's solve the first equation in terms of b.
b – a = 10
Add a to both sides.
b = 10 + a
Now we can substitute 10 + a in for b in the second equation.
(10 + a) + a = 56
10 + 2_a_ = 56
Subtract 10 from both sides.
2_a_ = 46
Divide both sides by 2.
a = 23
b = 10 + a = 10 + 23 = 33
The smallest value of S is 23, and the largest value is 33. Thus, S consists of the following set:
23, 25, 27, 29, 31, 33
We are told that T is equal to the squares of the elements of S. Thus, T consists of the following numbers:
529, 625, 729, 841, 961, 1089
We are asked ultimately to find the median of T. The median of any set is the middle number. In this case, T has six elements, so there are actually two numbers in the middle: 729 and 841. To find the median, we will take the average of the two middle numbers. The average of two numbers is equal to their sum divided by two.
median of T = (729 + 841)/2 = 1570/2 = 785.
The answer is 785.
The median of a set of eight consecutive odd integers is 22. What is the largest integer in the set?
The median of a set of eight consecutive odd integers is 22. What is the largest integer in the set?
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The set consists of eight consecutive odd integers. Let's call the integers, in order from least to greatest, A, B, C, D, E, F, G, and H. We are told that the median of the set is 22. The median in a set of numbers is the number in the middle. In the set of A, B, C, D, E, F, G, and H, the numbers represented by D and E would be in the middle. When we have two members in the middle, the median is equal to their average. Thus, 22 is equal to the average of D and E. In other words, 22 is the average of the 4th and 5th numbers in the set. Since 22 is halfway between 21 and 23, it makes sense that the 4th and 5th numbers would have to be 21 and 23, respectively. Thus, our set looks like this:
A, B, C, 21, 23, F, G, H
Beacuse the set consists of consecutive odd integers, each integer is two larger than the one before it. Therefore, this is the set of integers:
15, 17, 19, 21, 23, 25, 27, 29
The problem asks us to find the largest integer in the set, which is 29.
The answer is 29.
The set consists of eight consecutive odd integers. Let's call the integers, in order from least to greatest, A, B, C, D, E, F, G, and H. We are told that the median of the set is 22. The median in a set of numbers is the number in the middle. In the set of A, B, C, D, E, F, G, and H, the numbers represented by D and E would be in the middle. When we have two members in the middle, the median is equal to their average. Thus, 22 is equal to the average of D and E. In other words, 22 is the average of the 4th and 5th numbers in the set. Since 22 is halfway between 21 and 23, it makes sense that the 4th and 5th numbers would have to be 21 and 23, respectively. Thus, our set looks like this:
A, B, C, 21, 23, F, G, H
Beacuse the set consists of consecutive odd integers, each integer is two larger than the one before it. Therefore, this is the set of integers:
15, 17, 19, 21, 23, 25, 27, 29
The problem asks us to find the largest integer in the set, which is 29.
The answer is 29.
Find the median
9,6,1,6,5,9,8,3
Find the median
9,6,1,6,5,9,8,3
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To find the median, arrange the numbers from lowest to highest then find the middle number.
9,6,1,6,5,9,8,3
1,3,5,6,6,8,9,9
There are two numbers in the middle in this set (6 and 6).
In this case, the median is 6 but you would typically find the average of the two numbers in the middle.
To find the median, arrange the numbers from lowest to highest then find the middle number.
9,6,1,6,5,9,8,3
1,3,5,6,6,8,9,9
There are two numbers in the middle in this set (6 and 6).
In this case, the median is 6 but you would typically find the average of the two numbers in the middle.
x is the median of this set of numbers: x, 7, 12, 15, 19, 20, 8. What is one possible value of x?
x is the median of this set of numbers: x, 7, 12, 15, 19, 20, 8. What is one possible value of x?
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The median is the "middle number" in a sorted list of number; therefore, reorder the numbers in numerical order: 7, 8, 12, 15, 19, 20. Since x is the middle number, it must come between 12 and 15, leaving 14 as the only viable choice.
The median is the "middle number" in a sorted list of number; therefore, reorder the numbers in numerical order: 7, 8, 12, 15, 19, 20. Since x is the middle number, it must come between 12 and 15, leaving 14 as the only viable choice.
Craig has a jar full of loose change. He has 20 quarters, 15 dimes, 35 nickels and 55 pennies. If he orders them all from least to most valuable, what is the value of the median coin?
Craig has a jar full of loose change. He has 20 quarters, 15 dimes, 35 nickels and 55 pennies. If he orders them all from least to most valuable, what is the value of the median coin?
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The median is the coin which has an equal number of coins of lesser or equal value and greater or equal value on either side of it. The median therefore falls to one of the nickels (62 quarters, dimes and nickels above it and 62 nickels and pennies below it). The value of a nickel is 5 cents.
The median is the coin which has an equal number of coins of lesser or equal value and greater or equal value on either side of it. The median therefore falls to one of the nickels (62 quarters, dimes and nickels above it and 62 nickels and pennies below it). The value of a nickel is 5 cents.
A student decides to review all of his old tests from his math class in order to prepare for the upcoming exam. The student decides that any exam in which he received a 94% or better he should be able to skip since he understood the material well, but he will review the rest. The first 7 tests are scored out of 50 points. The rest are scored out of 40 points. In order, they got the following raw scores:
The student has 1 more test to add to this list. If he puts together all the tests he must review in order of percentile score, this final test is the median score. Which of the following cannot be the raw score of this final test?
A student decides to review all of his old tests from his math class in order to prepare for the upcoming exam. The student decides that any exam in which he received a 94% or better he should be able to skip since he understood the material well, but he will review the rest. The first 7 tests are scored out of 50 points. The rest are scored out of 40 points. In order, they got the following raw scores:
The student has 1 more test to add to this list. If he puts together all the tests he must review in order of percentile score, this final test is the median score. Which of the following cannot be the raw score of this final test?
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We first remove any 94%+ scores. For tests scored out of 50, a 94% is a 47. Thus anything 47 or greater will be removed. We see that this leaves us with:
For the second half, we have tests scored out of 40. 94% of 40 gives us 37.6, thus we will only remove tests that were a 38 or better. This leaves us with:
Now we have to convert them to percentiles. We end up with the lists as follows. The raw scores are in parentheses:
We then order it by percentile (removing raw scores, as they are unnecessary):
We notice that the median must be between the 78 and the 87.5 percentile scores. 31 out of 40 gives us 77.5%, which is below the score associated with an earlier test. Thus this cannot be the answer. The remaining 4 scores are all higher than 78%, and the largest of them is equal to 87.5 (the highest bound).
We first remove any 94%+ scores. For tests scored out of 50, a 94% is a 47. Thus anything 47 or greater will be removed. We see that this leaves us with:
For the second half, we have tests scored out of 40. 94% of 40 gives us 37.6, thus we will only remove tests that were a 38 or better. This leaves us with:
Now we have to convert them to percentiles. We end up with the lists as follows. The raw scores are in parentheses:
We then order it by percentile (removing raw scores, as they are unnecessary):
We notice that the median must be between the 78 and the 87.5 percentile scores. 31 out of 40 gives us 77.5%, which is below the score associated with an earlier test. Thus this cannot be the answer. The remaining 4 scores are all higher than 78%, and the largest of them is equal to 87.5 (the highest bound).
In the above set, which of the following is equal to the mean minus the median?
In the above set, which of the following is equal to the mean minus the median?
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The median is the middle value in an ordered set of numbers. In this instance, we have 8 numbers, so the median is the average of the 4th and 5th number:
The median for this set is 3.5. Now let's find the mean, or the average, of the set:
Now that we have the mean and median, we can finish the question. The question asks for the mean (3.375) minus the median (3.5):
The median is the middle value in an ordered set of numbers. In this instance, we have 8 numbers, so the median is the average of the 4th and 5th number:
The median for this set is 3.5. Now let's find the mean, or the average, of the set:
Now that we have the mean and median, we can finish the question. The question asks for the mean (3.375) minus the median (3.5):