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# Statistics

1. Explain following concepts:

- distribution characteristics (average)
- variability characteristics

Solution:

b) Geometric mean:

c) Harmonic mean:

d) Mode mod(x) is the value that occurs the most frequently in a data set or a probability distribution.

e) Median med(x) is:

R = x

b) Variance (dispersion)

c) Standard deviation

## Distribution characteristics

a) Arithmetic mean:b) Geometric mean:

c) Harmonic mean:

d) Mode mod(x) is the value that occurs the most frequently in a data set or a probability distribution.

e) Median med(x) is:

- the middle value of a data set sorted by the value containing odd number of elements
- aritmethmetic mean of the two middle values of a data set sorted by the value containing even number of elements

## Variability characteristics

a) RangeR = x

_{max}- x_{min}b) Variance (dispersion)

c) Standard deviation

2. Following values were measured in the lab (in milimeters):

{302;310;312;310;313;318;305;309;310;309}

Determine the arithmetic mean, geometric mean, geometrický priemer, mode and median.

Solution:

Sorted values:

{302;305;309;309;310;310;310;312;313;318}

Sorted values:

{302;305;309;309;310;310;310;312;313;318}

3.A car hit an average speed of v_{1} = 20 kmph in the first half of the journey, and hit an average speed of v_{2} = 80 kmph in the second half of the journey. Determine the average speed of the car.

Solution:

The car moved in different speeds, therefore it passed each half in different time.

The average speed equals to the harmonic mean of v

The average speed of the car was 32 kmph.

The car moved in different speeds, therefore it passed each half in different time.

The average speed equals to the harmonic mean of v

_{1}and v_{2}.The average speed of the car was 32 kmph.

4.Following values were measured to determine the weight of a cog wheel. Find out the interval which contains the real weight value with a probability of 68%.

M = {324;330;327;319;334;304}

Solution:

5. Hounters A and B compete in target shooting. Who was more accurate and won the competition?

Results:

A = {9;8;8;8;7}

B = {10;10;8;7;5}

A = {9;8;8;8;7}

B = {10;10;8;7;5}

Solution:

Hunter A

Hunter B

Variance of hunter A equals s

Hunter A won the competition.

Hunter A

Hunter B

Variance of hunter A equals s

^{2}(A) = 0,4 , variance of hunter B equals s^{2}(B) = 3,6. Stands s^{2}(A) < s^{2}(B).Hunter A won the competition.

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