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2 years ago

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Mathematics

1 answer:

leva [86]2 years ago

6 0

Using it's definitions, the five-number summary and the interquartile range for the data-set is given as follows:

Minimum: 6

Lower quartile: 8

Median: 21.

Upper quartile: 27

Maximum: 35

Interquartile range: 19

<h3>What are the median and the quartiles of a data-set? How to find the interquartile range using it?</h3>

The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile.

The first quartile is the median of the first half of the data-set.

The third quartile is the median of the second half of the data-set.

The interquartile range is the difference of the third quartile and the first quartile.

This data-set has 12 elements, which is an even number, hence the median is the mean of the 6th and 7th elements, as follows:

Me = (19 + 23)/2 = 21.

The minimum is the lowest value in the data-set, which is of 6, while the maximum is of 35, which is the largest value in the data-set.

The first quartile is the median of the first half, composed by 6, 7, 8, 8, 16, which is the third element of 8.

The third quartile is the median of the second half, composed by 23, 24, 26, 27, 34, 35, which is of 27. Hence the interquartile range is of 27 - 8 = 19.

More can be learned about the five-number summary and the interquartile range at brainly.com/question/3876456

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Answer:

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Step-by-step explanation:

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3 years ago

QUESTION 11.1

lawyer [7]

The group paid $ 5250 at first city and $ 6250 at second city

<u>Solution:</u>

Let x = the charge in 1st city before taxes

Let y = the charge in 2nd city before taxes

The hotel charge before tax in the second city was $1000 higher than in the first

Then the charge at the second hotel before tax will be x + 1000

y = x + 1000 ----- eqn 1

The tax in the first city was 8.5% and the tax in the second city was 5.5%

The total hotel tax paid for the two cities was $790

<em><u>Therefore, a equation is framed as:</u></em>

8.5 % of x + 5.5 % of y = 790

$\frac{8.5}{100} \times x + \frac{5.5}{100} \times y = 790$

0.085x + 0.055y = 790 ------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

<em><u>Substitute eqn 1 in eqn 2</u></em>

0.085x + 0.055(x + 1000) = 790

0.085x + 0.055x + 55 = 790

0.14x = 790 - 55

0.14x = 735

<em><u>Substitute x = 5250 in eqn 1</u></em>

y = 5250 + 1000

Thus the group paid $ 5250 at first city and $ 6250 at second city

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3 years ago

The table showing the stock price changes for a sample of 12 companies on a day is contained in the Excel file below.

AfilCa [17]

Answer:

(a) The sample variance for the daily price change is 0.2501.

(b) The sample standard deviation for the daily price change is 0.5001.

Step-by-step explanation:

Let the random variable <em>X</em> denote the stock price changes for a sample of 12 companies on a day.

The data provided is:

<em>X</em> = {0.82 , 1.44 , -0.07 , 0.41 , 0.21 , 1.33 , 0.97 , 0.30 , 0.14 , 0.12 , 0.42 , 0.15}

(a)

The formula to compute the sample variance for the daily price change is:

$s^{2}=\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}$

The sample mean is computed using the formula:

$\bar X=\frac{1}{n}\sum\limits^{12}_{i=1}{X_{i}}$

Consider the Excel output attached below.

In Excel the formula to compute the sample mean and sample variance are:

$\bar X$ =AVERAGE(A2:A13)

$s^{2}$ =VAR.S(A2:A13)

Thus, the sample variance for the daily price change is 0.2501.

(b)

The formula to compute the sample standard deviation for the daily price change is:

$s=\sqrt{\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}}$

Consider the Excel output attached below.

In Excel the formula to compute the sample standard deviation is:

$s$ =STDEV.S(A2:A13)

Thus, the sample standard deviation for the daily price change is 0.5001.

(c)

The (1 - <em>α</em>)% confidence interval for population variance is:

$CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]$

Compute the critical value of Chi-square for <em>α</em> = 0.05 and (n - 1) = (12 - 1) = 11 degrees of freedom as follows:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2,11}=21.920$

$\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{(1-0.05/2),11}=\chi^{2}_{0.975,11}=3.816$

*Use a Chi-square table.

Compute the 95% confidence interval estimates of the population variance as follows:

$CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]$

$=[\frac{(12-1)\times 0.2501}{21.920 } \leq \sigma^{2}\leq \frac{(12-1)\times 0.2501}{3.816} ]$

$=[0.125506\leq \sigma^{2}\leq 0.720938]\\\approx [0.1255, 0.7210]$

Thus, the 95% confidence interval estimates of the population variance is (0.1255, 0.7210).

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4 years ago