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4 years ago

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fred thinks of a number that is greater than 110 and less than 120. he can make his number with same number of hundreds,tens and

ones. what is freds number?

1 answer:

VashaNatasha [74]4 years ago

8 0

The number Fred thinks of is 111

<em><u>Solution:</u></em>

Given that, Fred thinks of a number that is greater than 110 and less than 120

He can make his number with same number of hundreds,tens and ones

<em><u>To find: Fred's number</u></em>

Given that the number is greater than 110 and less than 120

The possible numbers are:

111 , 112, 113, 114, 115, 116, 117, 118, 119

But he can make his number with same number of hundreds,tens and ones

So the numbers in ones, tens and hundreds place are same

So the only possible number is 111

Thus the number Fred thinks of is 111

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Calculate the sample standard deviation and sample variance for the following frequency distribution of heart rates for a sample

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Answer:

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$\sigma^2 = 157.2$ ---- sample variance

Step-by-step explanation:

Solving (a): The sample variance

First, we calculate the midpoint of each class (this is the average of the limits)

So, we have:

$x_1 = \frac{56 + 64}{2} = 60$

$x_2 = \frac{65 + 73}{2} = 69$

And so on

So, the table becomes:

$\begin{array}{cc}{x} & {f} & {60} & {11} & {69} & {15} & {78} & {14} & {87} & {4} & {96} & 11 \ \end{array}$

Calculate the mean

$\bar x = \frac{\sum fx}{\sum f}$

$\bar x = \frac{60*11+69*15+78*14+87*4+96*11}{11+15+14+4+11}$

$\bar x = \frac{4191}{55}$

$\bar x = 76.2$

The variance is:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f - 1}$

So, we have:

$\sigma^2 = \frac{(60 - 76.2)^2*11+(69 - 76.2)^2*15+(78 - 76.2)^2 *14+(87 - 76.2)^2*4+(96 - 76.2)^2*11}{11+15+14+4+11-1}$

$\sigma^2 = \frac{8488.8}{54}$

$\sigma^2 = 157.2$

The sample standard deviation is:

$\sigma = \sqrt{\sigma^2$

$\sigma = \sqrt{157.2$

$\sigma = 12.5$

Solving (b): The sample variance

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$\sigma^2 = 157.2$

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