STA301 GDB Fall 2020 Solution & Discussion

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STA301 GDB Fall 2020 Solution & Discussion

skewed okay subsequently symmetrical distribution

in the case of symmetrical distribution

the empirical relation state that

those symmetrical distributions

mean

when our distribution is not symmetric

mean is greater than median and median

is greater than

mod sithara negative skewed

the empirical relation is mean is

less than median and median is less than

more according to carl pearson

in case of moderately skewed

or moderately asymmetrical distribution

the value of the mean

median and mode have the

following empirical relations in case of

relation

mean minus mode is equal to

3 into a negative

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STA301 GDB Solution Fall 2020 

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STA301 GDB Solution Fall 2020 26-11-2020

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STA301 GDB Fall 2020 100% Correct Solution

Description skewed okay subsequently symmetrical distribution in the case of symmetrical distribution the empirical relation state that those symmetrical distributions mean when our distribution is not symmetric mean is greater than median and median is greater than mod sithara negative skewed the empirical relation is mean is less than median and median is less than more according to carl pearson in case of moderately skewed or moderately asymmetrical distribution the value of the mean median and mode have the following empirical relations in case of relation mean minus mode is equal to 3 into a negative

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STA301 GDB 1 Solution 2020 By Maria Parveen || STA301 GDB 1 Solution Fall 2020

STA01 GDB - 1 Solution fall 2020 ll VU Learning STA101 GDB 1 Solution fall 2020 STA301 GDB - 1 Solution fall 2020 STA301 GDB 1 Solution 2020

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1)

Mode=3Median —2 Mean

32_1=3 Median-2(354) 32.1=3

Median-70.8

Median=102.913

Median=34.3

Positive skewed distribution

2)

Mode=3

Median - 2

Mean Mode=3(1459)-2(1403)

Mode=4377-2806 Mode=1571

Negative skewed distribution

3)

Mode= 3 Median - 2Mean

50=3(50)-2 Mean 2

Meau=150-50 Mean=100/2 Mean=50

Symmetric skewed distribution

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STA301 GDB Fall 2020 Attachment  File

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STA-301

GDB No 1 Solution

2020 by Raise for Success

Requirement:

What can you say of the skewness in each of the following cases?

• Mode= 32.1 & Mean= 35.4
• Median= 1459 & Mean= 1403
• Median= 50 & Mean= 50

Answer:

Mode = 3Median - 2Mean

32.1 = 3Median – 2(35.4)

32.1 = 3Median – 70.8

32.1 + 70.8 = 3Median

102.9 = 3Median

102.9 / 3 = Median

34.3 = Median

Mean > Median > Mode

35.4 > 34.3 > 32.1 Positive skewed distribution

Mode = 3Median - 2Mean

Mode = 3(1459) – 2(1403)

Mode = 4377 – 2806

Mode = 1571

Mean < Median < Mode

1403 > 1459 > 1571 Negative skewed distribution

Mode = 3Median - 2Mean

50 = 3(50) – 2Mean

50 = 150 – 2Mean

50 - 150= - 2Mean

-100 = - 2Mean

100 / 2 = Mean

50 = Mean

Mean = Median = Mode

50 = 50 = 50 Normal distribution

 

 

[3pexf8bxro8ag]

 

STA-301

GDB No 1 Solution

2020 by Raise for Success

Requirement:

What can you say of the skewness in each of the following cases?

• Mode= 32.1 & Mean= 35.4
• Median= 1459 & Mean= 1403
• Median= 50 & Mean= 50

Answer:

Mode = 3Median - 2Mean

32.1 = 3Median – 2(35.4)

32.1 = 3Median – 70.8

32.1 + 70.8 = 3Median

102.9 = 3Median

102.9 / 3 = Median

34.3 = Median

Mean > Median > Mode

35.4 > 34.3 > 32.1 Positive skewed distribution

Mode = 3Median - 2Mean

Mode = 3(1459) – 2(1403)

Mode = 4377 – 2806

Mode = 1571

Mean < Median < Mode

1403 > 1459 > 1571 Negative skewed distribution

Mode = 3Median - 2Mean

50 = 3(50) – 2Mean

50 = 150 – 2Mean

50 - 150= - 2Mean

-100 = - 2Mean

100 / 2 = Mean

50 = Mean

Mean = Median = Mode

50 = 50 = 50 Normal distribution

 

 

[3pexf8bxro8ag]

 

STA-301

GDB No 1 Solution

Requirement:

What can you say of the in each of the following cases?

• Mode= 32.1 & Mean= 35.4
• Median= 1459 & Mean= 1403
• Median= 50 & Mean= 50

Answer:

Mode = 3Median - 2Mean

32.1 = 3Median – 2(35.4)

32.1 = 3Median – 70.8

32.1 + 70.8 = 3Median

102.9 = 3Median

102.9 / 3 = Median

34.3 = Median

Mean > Median > Mode

35.4 > 34.3 > 32.1 Positive skewed distribution

Mode = 3Median - 2Mean

Mode = 3(1459) – 2(1403)

Mode = 4377 – 2806

Mode = 1571

Mean < Median < Mode

1403 > 1459 > 1571 Negative skewed distribution

Mode = 3Median - 2Mean

50 = 3(50) – 2Mean

50 = 150 – 2Mean

50 - 150= - 2Mean

-100 = - 2Mean

100 / 2 = Mean

50 = Mean

Mean = Median = Mode

50 = 50 = 50 Normal distribution

 

 

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STA301 GDB Fall 2020

STA-301

GDB No 1 Solution

Requirement:

What can you say of the in each of the following cases?

• Mode= 32.1 & Mean= 35.4
• Median= 1459 & Mean= 1403
• Median= 50 & Mean= 50

Answer:

Mode = 3Median - 2Mean

32.1 = 3Median – 2(35.4)

32.1 = 3Median – 70.8

32.1 + 70.8 = 3Median

102.9 = 3Median

102.9 / 3 = Median

34.3 = Median

Mean > Median > Mode

35.4 > 34.3 > 32.1 Positive skewed distribution

Mode = 3Median - 2Mean

Mode = 3(1459) – 2(1403)

Mode = 4377 – 2806

Mode = 1571

Mean < Median < Mode

1403 > 1459 > 1571 Negative skewed distribution

Mode = 3Median - 2Mean

50 = 3(50) – 2Mean

50 = 150 – 2Mean

50 - 150= - 2Mean

-100 = - 2Mean

100 / 2 = Mean

50 = Mean

Mean = Median = Mode

50 = 50 = 50 Normal distribution

 

 

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STA301 Statistics and Probability GDB 1 Solution & Discussion Fall 2020

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STA301 GDB Solution:

 

 

Q= What can you say of the Skewness in each of following cases?

1= Mode= 32.1   and Mean= 35.4

2= Median= 1459 and Mean= 1403

3= Median= 50 and Mode= 50

 

Answer:

They ask us find about Skewness: So there are two types of Skewed.

1= Positive Skewed: Mean<Median<Mode

2= Negative Skewed: Mode>Median>Mean

Solution.

1= Mode= 32.1   and Mean= 35.4  So we find Median= ?

Formula: Mode=3 Medain-2 Mean

32.1= 3(Median) -2(35.4)

32.1= 3(Median) -70.8

32.1+70.8= 3(Median)

102.9= 3(Median)

102.9/3 = Median

34.3= Median

So it is Positive Skewed : Mean<Median<Mode = 35.4<34.3<32.1

 

2= Median= 1459 and Mean= 1403 So we find Mode=?

Formula: Mode = 3Median -2Mean

Mode = 3Medain -2Mean

Mode= 3(1459) -2(1403)

Mode= 4377-2806

Mode= 1571

So this is Negative Skewed: Mode>Median>Mean = 1571>1459>1403

 

3= Median= 50 and Mode= 50 So we Find Mean=?

Fomula: Mode = 3Medain -2Mean

Mode = 3Medain -2Mean

50= 3(50) -2(Mean)

2(Mean)= 150-50

2(Mean)= 100

2(Mean)/2 = 100/2 = 50

So, Mean= 50

So This is Normal Distribution: Mean=Median=Mode: 50=50=50