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# Assignment For STA301 Statistics and Probability,Semester Fall 2015 (Due date:Dec 07, 2015)

Assignment no.1 (Lessons 1 – 15)

Question 1:                                                                                                                Marks: 3+2=5

a)      The mean age of a group of 100 persons was found to be 32.02. Later it was discovered that age 57 was misread as 27. Find the corrected mean?

b)      The sum of deviations from an arbitrary value A = 15 for 10 values is 25, find the A.M.

Question 2:                                                                                                               Marks: 3+3=6

a)      The mean of 200 items is 48 and their S.D is 3. Find the values of  and

b)      A vehicle going up from Islamabad to Murree consumes petrol at the rate of One liter per 8 km, while going down from Murree to Islamabad one liter per 12 km. Find the average rate of consumption between the two cities.

Question 3:                                                                                                                         Marks: 4

The S.D of the symmetrical distribution is 5. What must be the value of, and also explain that the distribution will be?

(i)                 Meso-kurtic

(ii)               Platy-Kurtic

(iii)             Leptokurtic

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Question 1: b)      The sum of deviations from an arbitrary value A = 15 for 10 values is 25, find the A.M.

Fall 2015_STA301_ 1ST Assignment  Solution

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Mean of 200 items is 48 and their standard deviation is 3. Find the sum of all the items and the sum of the squares of all the items.

Solution:

Mean of 100 items (x̄) = 48

Standard deviation (σ) = 3

here we have to answer for two questions that is sum of all items (Σ x) and sum of squares all items (Σ x²).

x̄ = (Σ x/n)

48 = (Σ x/200)

Σ x = 200 x 48

= 9600

Sum of all items = 9600

To find sum of squares of all items, we have to find variance (σ²).

σ² = (Σ x²/n) - (Σ x/n)²

(3)² =  (Σ x²/200) - (9600/200)²

(3)² =  (Σ x²/200) - (48)²

9 = (Σ x²/200) - 2304

9 + 2304 = (Σ x²/200)

2313 = Σ x²/200

Σ x² = 2313(200)

Σ x² = 462600

Sum of squares of all items = 462600

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