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max2010maxim [7]
2 years ago
12

If the mean of 5 positive integers is 15, what is the maximum possible difference between the largest and the smallest of these

5 numbers?
Mathematics
1 answer:
igomit [66]2 years ago
4 0

Answer:

64

Step-by-step explanation:

If the mean is 15, the sum of 5 numbers is:

  • 5*15 = 75

Minimum value for the first four numbers would be:

  • 1, 2, 3, 4

Then the fifth number is:

  • 75 - (1+2+3+4) = 75 - 10 = 65

So the maximum difference is:

  • 65 - 1 = 64
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Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a
spin [16.1K]

Answer:

Mean = 68.9

s^2 =18.1 --- Variance

Step-by-step explanation:

Given

\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5

For 59 - 66

x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5

For 67 - 74

x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5

For 75 - 82

x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5

For 83 - 90

x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5

So, the table becomes:

\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}

The mean is then calculated as:

Mean = \frac{\sum fx}{\sum f}

Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}

Mean = \frac{2547.5}{37}

Mean = 68.9 -- approximated

Solving (b): The sample variance:

This is calculated as:

s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}

So, we have:

s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}

s^2 =\frac{652.8}{36}

s^2 =18.1 -- approximated

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alekssr [168]

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Answer:

C & D

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