If the points of inflexion of a normal curve are 40 and 60 respectivel...
Given: Points of inflexion of a normal curve are 40 and 60 respectively.
To find: Mean deviation.
Solution:
We know that the points of inflexion of a normal curve are symmetric about the mean.
Let the mean of the normal curve be 'M'.
So, the distance of the points of inflexion from the mean would be:
(60 - M) = (M - 40)
Solving this equation, we get:
M = 50
So, the mean of the normal curve is 50.
Now, we need to find the mean deviation.
Mean deviation is given by the formula:
Mean deviation = ∫|x - M| f(x) dx / ∫f(x) dx
Where, f(x) is the probability density function of the normal curve.
For a standard normal curve, the mean deviation is 1.
But in this case, we need to adjust the formula for the mean deviation based on the given mean 'M'.
So, the adjusted formula for mean deviation is:
Mean deviation = ∫|x - 50| f(x) dx / ∫f(x) dx
Now, we can use the properties of the normal curve to simplify this formula.
We know that the normal curve is symmetric about the mean.
So, we can split the integral into two parts:
∫|x - 50| f(x) dx = ∫(x - 50) f(x) dx for x > 50
+ ∫(50 - x) f(x) dx for x < />
Since the normal curve is a continuous probability distribution, we can assume that the probability density function is continuous and differentiable.
So, we can use the fact that the derivative of the normal curve is proportional to the distance from the mean.
So, we have:
f'(x) = k (x - 50) f(x) for x > 50
f'(x) = -k (x - 50) f(x) for x < />
Where, k is a constant of proportionality.
Integrating both sides, we get:
ln(f(x)) = k/2 (x - 50)^2 + C for x > 50
ln(f(x)) = -k/2 (x - 50)^2 + C for x < />
Where, C is a constant of integration.
Taking the exponential of both sides, we get:
f(x) = A exp(k/2 (x - 50)^2) for x > 50
f(x) = A exp(-k/2 (x - 50)^2) for x < />
Where, A is a constant of proportionality.
Now, we can use these formulas to evaluate the integrals.
∫(x - 50) f(x) dx for x > 50
= ∫(x - 50) A exp(k/2 (x - 50)^2) dx for x > 50
= A/k exp(k/2 (x - 50)^2) + C1 for x > 50
Where, C1 is a constant of integration.
∫(50 - x) f(x) dx for x < />
= ∫(50 - x) A exp(-k/2 (x - 50)^2) dx for x < />
If the points of inflexion of a normal curve are 40 and 60 respectivel...
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