Question #22307

For a Normal distribution with mean, μ=2, and standard deviation, σ=4, 30% of all observations have a value less than
For a Normal distribution with mean, μ=2, and standard deviation, σ=4, 40% of all observations have a value greater than
For a normal distribution with mean, μ=2, and standard deviation, σ=4 what proportion of observations take values less than 1?
Can you also please explain how to get the answer?

Expert's answer

Let X be random variable having standard normaldistribution, that ismean μ=0, and

standard deviation, σ=1.

Let also F be the cumulative distribution function for X,so

F(t) = P(X<t)

The values of function F are computed and can be found inany book on probability theory.

So for problems 1)-3) we will use the values of functionF.

Let Y be the random variable having Normal distributionwith mean, μ=2, and standard deviation, σ=4.In 1) and 2) we should find values y1, y2 such that

1) P(Y < y1) = 0.32) P(Y > y2) = 0.4

In 3) we should find probability

3) P(Y < 1).

Consider the so called "normalized" randomvariable Z = (Y-μ)/σ =(Y-2)/4.

Then Z has standard normal distribution and so

P(Z<t) = F(t).

Now we can solve 1)-3).1)

0.3 = P(Y<y1) = P((Y-2)/4 < (y1-2)/4 )

= P( Z <(y1-2)/4 )

= F( (y1-2)/4)

From tables for F we obtain that F(-0.524) = 0.3

whence

(y1-2)/4 = -0.542

Therefore

y1 = -0.542*4+2=-0.168

Thus for a Normal distribution with mean, μ=2, andstandard deviation, σ=4,

30% of all observations have a value less than -0.1682)

0.4 = P(Y>y2)

= P((Y-2)/4 > (y1-2)/4 )

= P( Z >(y2-2)/4 )

= 1 - P( Z< (y2-2)/4 )

= 1 - F((y2-2)/4 )

whence F( (y2-2)/4 ) =1-0.4=0.6.

From tables for Fwe obtain that

F(0.253) = 0.6

whence

(y2-2)/4 = 0.253

Therefore

y1 = 0.253*4+2=3.012

Thus for a Normal distribution with mean, μ=2, andstandard deviation, σ=4,

40% of all observations have a value greater than 3.0123)

P(Y < 1) = P((Y-2)/4 < (1-2)/4 )

= P( Z< -1/4 )

= P( Z< -0.25 )

=F(-0.25)

= 0.401

Thus for a normal distribution with mean, μ=2, andstandard deviation, σ=4

40.1 % of observations take values less than 1

standard deviation, σ=1.

Let also F be the cumulative distribution function for X,so

F(t) = P(X<t)

The values of function F are computed and can be found inany book on probability theory.

So for problems 1)-3) we will use the values of functionF.

Let Y be the random variable having Normal distributionwith mean, μ=2, and standard deviation, σ=4.In 1) and 2) we should find values y1, y2 such that

1) P(Y < y1) = 0.32) P(Y > y2) = 0.4

In 3) we should find probability

3) P(Y < 1).

Consider the so called "normalized" randomvariable Z = (Y-μ)/σ =(Y-2)/4.

Then Z has standard normal distribution and so

P(Z<t) = F(t).

Now we can solve 1)-3).1)

0.3 = P(Y<y1) = P((Y-2)/4 < (y1-2)/4 )

= P( Z <(y1-2)/4 )

= F( (y1-2)/4)

From tables for F we obtain that F(-0.524) = 0.3

whence

(y1-2)/4 = -0.542

Therefore

y1 = -0.542*4+2=-0.168

Thus for a Normal distribution with mean, μ=2, andstandard deviation, σ=4,

30% of all observations have a value less than -0.1682)

0.4 = P(Y>y2)

= P((Y-2)/4 > (y1-2)/4 )

= P( Z >(y2-2)/4 )

= 1 - P( Z< (y2-2)/4 )

= 1 - F((y2-2)/4 )

whence F( (y2-2)/4 ) =1-0.4=0.6.

From tables for Fwe obtain that

F(0.253) = 0.6

whence

(y2-2)/4 = 0.253

Therefore

y1 = 0.253*4+2=3.012

Thus for a Normal distribution with mean, μ=2, andstandard deviation, σ=4,

40% of all observations have a value greater than 3.0123)

P(Y < 1) = P((Y-2)/4 < (1-2)/4 )

= P( Z< -1/4 )

= P( Z< -0.25 )

=F(-0.25)

= 0.401

Thus for a normal distribution with mean, μ=2, andstandard deviation, σ=4

40.1 % of observations take values less than 1

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