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STA301 GDB Fall 2020 Solution & Discussion

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

STA301 GDB Solution Fall 2020 26-11-2020

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

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

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

STA301 GDB Fall 2020 Attachment  File

Attachments:

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

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

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

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

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

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

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

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

STA301 Statistics and Probability GDB 1 Solution & Discussion Fall 2020

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

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

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