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leva [86]
2 years ago
6

What is the interquartile range of the following data set? 45,12,48,96,61,84,29,1,72,5,14

Mathematics
2 answers:
nadezda [96]2 years ago
7 0

Answer: 60

Step-by-step explanation: Apex said so

slava [35]2 years ago
6 0

Answer:

60

Step-by-step explanation:

Population size : 11

To calculate inter-quartile range

  • Order the data from least to greatest.
  • Find the median.
  • Calculate the median of both the lower and upper half of the data.
  • The IQR is the difference between the upper and lower medians.

Step 1

1, 5 , 12 ,14 , 29 , 45 , 48, 61 , 72 , 84 , 96

Step 2

medium = 45

Step 3

Q1

1 , 5 , 12 ,14 , 29

median = 12

Q3

48 , 61 , 72 , 84 , 96

median = 72

Step 4

IQ = Q3 - Q1

    = 72 - 12

    = 60

You might be interested in
Lenovo uses the​ zx-81 chip in some of its laptop computers. the prices for the chip during the last 12 months were as​ follows:
Stella [2.4K]
Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months

\begin{tabular}
{|c|c|c|c|}
Month&Price per Chip&Month&Price per Chip\\[1ex]
January&\$1.90&July&\$1.80\\
February&\$1.61&August&\$1.83\\
March&\$1.60&September&\$1.60\\
April&\$1.85&October&\$1.57\\
May&\$1.90&November&\$1.62\\
June&\$1.95&December&\$1.75
\end{tabular}

The forcast for a period F_{t+1} is given by the formular

F_{t+1}=\alpha A_t+(1-\alpha)F_t

where A_t is the actual value for the preceding period and F_t is the forcast for the preceding period.

Part 1A:
Given <span>α ​= 0.1 and the initial forecast for october of ​$1.83, the actual value for october is $1.57.

Thus, the forecast for period 11 is given by:

F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\  \\ =0.1(1.57)+(1-0.1)(1.83) \\  \\ =0.157+0.9(1.83)=0.157+1.647 \\  \\ =1.804

Therefore, the foreast for period 11 is $1.80


Part 1B:

</span>Given <span>α ​= 0.1 and the forecast for november of ​$1.80, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

F_{12}=\alpha&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.1(1.62)+(1-0.1)(1.80) \\  \\ &#10;=0.162+0.9(1.80)=0.162+1.62 \\  \\ =1.782

Therefore, the foreast for period 12 is $1.78</span>



Part 2A:

Given <span>α ​= 0.3 and the initial forecast for october of ​$1.76, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

F_{11}=\alpha&#10; A_{10}+(1-\alpha)F_{10} \\  \\ =0.3(1.57)+(1-0.3)(1.76) \\  \\ &#10;=0.471+0.7(1.76)=0.471+1.232 \\  \\ =1.703

Therefore, the foreast for period 11 is $1.70

</span>
<span><span>Part 2B:

</span>Given <span>α ​= 0.3 and the forecast for November of ​$1.70, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

F_{12}=\alpha&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.3(1.62)+(1-0.3)(1.70) \\  \\ &#10;=0.486+0.7(1.70)=0.486+1.19 \\  \\ =1.676

Therefore, the foreast for period 12 is $1.68



</span></span>
<span>Part 3A:

Given <span>α ​= 0.5 and the initial forecast for october of ​$1.72, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

F_{11}=\alpha&#10; A_{10}+(1-\alpha)F_{10} \\  \\ =0.5(1.57)+(1-0.5)(1.72) \\  \\ &#10;=0.785+0.5(1.72)=0.785+0.86 \\  \\ =1.645

Therefore, the forecast for period 11 is $1.65

</span>
<span><span>Part 3B:

</span>Given <span>α ​= 0.5 and the forecast for November of ​$1.65, the actual value for November is $1.62

Thus, the forecast for period 12 is given by:

F_{12}=\alpha&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.5(1.62)+(1-0.5)(1.65) \\  \\ &#10;=0.81+0.5(1.65)=0.81+0.825 \\  \\ =1.635

Therefore, the forecast for period 12 is $1.64



Part 4:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span></span></span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78

Thus, the mean absolute deviation is given by:

\frac{|1.57-1.83|+|1.62-1.80|+|1.75-1.78|}{3} = \frac{|-0.26|+|-0.18|+|-0.03|}{3}  \\  \\ = \frac{0.26+0.18+0.03}{3} = \frac{0.47}{3} \approx0.16

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α ​= 0.1 of October, November and December is given by: 0.157



</span><span><span>Part 5:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68

Thus, the mean absolute deviation is given by:

&#10; \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = &#10;\frac{|-0.17|+|-0.08|+|-0.07|}{3}  \\  \\ = \frac{0.17+0.08+0.07}{3} = &#10;\frac{0.32}{3} \approx0.107

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α ​= 0.3 of October, November and December is given by: 0.107



</span></span>
<span><span>Part 6:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span><span>α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64

Thus, the mean absolute deviation is given by:

&#10; \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = &#10;\frac{|-0.15|+|-0.03|+|0.11|}{3}  \\  \\ = \frac{0.15+0.03+0.11}{3} = &#10;\frac{29}{3} \approx0.097

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α ​= 0.5 of October, November and December is given by: 0.097</span></span>
5 0
1 year ago
The random variable X is exponentially distributed, where X represents the waiting time to see a shooting star during a meteor s
faltersainse [42]

Answer:

The parameters of the exponential distribution is 0.0133.

Step-by-step explanation:

Exponential distribution is a continuous probability distribution.

The density function of exponential distribution is,  

f _{X}(x)=\theta \cdot e^{-\theta\cdot x},\ x\geq 0

Here the parameter θ is the reciprocal of the mean of the random variable <em>X</em>.

The random variable <em>X</em> has an average value of 75 seconds.

Compute the parameters of the exponential distribution as follows:

\theta=\frac{1}{\mu}\\\\

  =\frac{1}{75}\\\\=0.013333333333\\\\\approx 0.0133

Thus, the parameters of the exponential distribution is 0.0133.

3 0
1 year ago
A bike wheel is 26 inches in diameter. What is the bike wheel's diameter in millimeters (1 inch = 25.4 millimeters)?
ruslelena [56]

Answer:

Diameter of wheel in millimetres is 660.4

Step-by-step explanation:

Diameter of wheel in inches = 26

given

1 inch = 25.4 millimeters

multiplying RHS and LHS by 26

26*1 inch = 26*25.4 millimeters

=>26 inch = 660.4 mm.

Thus, diameter of wheel in millimetres is 660.4

3 0
2 years ago
Three friends are playing catch. Zoe is in a straight path 12 feet to the west of alex. Jin is in a straight path 9feet to the n
Tanzania [10]

Answer:

15 ft

Step-by-step explanation:

This problem can be represented by a right angle triangle, shown in the diagram below.

The distance between Jin and Zoe is the hypotenuse of the triangle, x.

According to Pythagoras theorem,

hyp² = opp² + adj²

Where opp is the opposite side and adjacent is the adjacent side to any angle of consideration (which is not important in this case)

Hence:

x² = 12² + 9²

x² = 144 + 81

x² = 225

Finding the square root:

x = 15 ft

Jin and Zoe are 15 ft apart.

4 0
2 years ago
A statue is mounted on top of a 16 foot hill. From the base of the hill to where you are standing is 77 feet and the statue subt
DanielleElmas [232]

Consider right triangle with vertices B - base of the hill, S - top of the statue and Y - you. In this triangle angle B is right and angle Y is 13.2°. If h is a height of the statue, then the legs YB and BS have lengths 77 ft and 16+h ft.

You have lengths of two legs and measure of one acute angle, then you can use tangent to find h:

\tan 13.2^{\circ}=\dfrac{\text{opposite leg}}{\text{adjacent leg}} =\dfrac{16+h}{77}, \\ \\ 0.2345=\dfrac{16+h}{77},\\  \\ 16+h=0.2345\cdot 77=18.0565,\\ h=18.0565-16=2.0565 ft.

Answer: the height of the statue is 2.0565 ft.

8 0
1 year ago
Read 2 more answers
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