Find the mean and SD of marks in an examination where 44% of the candidates obtained marks below 55 and 6% get above 80 marks
Answers
Answer:
ok i will do but can u please tell which class question is this
Answer:
mean = 61.578 and SD = 43.85
Step-by-step explanation:
Given:- 44% of the candidates obtained marks below 55 and 6% get above 80 marks.
To Find:- Mean and Standard deviation of marks.
Solution:-
x = 55 and x = 80.
i) Since 44% of items are under x = 55, position of x will be the left of the ordinate x = µ.
ii) Since 6% of items are above x = 80 , position of this x will be to the right of ordinate x = µ.
When x = 55, z = (x - µ)/σ = (55 - µ)/σ = - z1 (say)
Since x is left of x = µ , z1 is taken as negative.
When x = 80, z = (x - µ)/σ = (80 - µ)/σ = z2 (say)
⇒ P(x < 55) = 0.44
⇒ P(z < - z1) = 0.44
⇒ P(- z1 < z < 0) = P(- ∞ < z < 0) – p(- ∞ < z < z1)
s = 0.5 - 0.44 = 0.06
⇒ P(0 < z < z1) = 0.06 (by symmetry)
z1 = 0.15
Also P(x > 80) = 0.06
P(0 < z < z2) = P(0 < z < ∞) – P(z2 < z < ∞)
= 0.5 - 0.06 = 0.44
z2 = 1.42
Substituting the values of z1 and z2, we get
(55 - µ)/σ = - 0.15 and (80 - µ)/σ = 1.42
Solving µ - 0.15σ = 55 ----------- ( 1 )
µ + 1.42σ = 80 ----------- ( 2 )
(2) - (1) ⇒ 0.57σ = 25 ⇒ σ = 43.85
Substituting σ = 43.85 in ( 1 ),
µ = 55 + 0.15 (43.85)
= 55 + 6.578
= 61.578
Therefore, mean = 61.578 and SD = 43.85
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