Answer to Question #127872 in Statistics and Probability for Sourabh

Question

suppose x is a normally distributed random variable with μ = 50 and ϭ = 3. find a value of the random variable, call it x0, such that a. p(x ≤ x0) = 0.8413 b. p(x > x0) = 0.25 c. p(x > x0) = 0.95 d. p(41 ≤ x < x0) = 0.8630 e. 10% of the values of x are less than x0. f. 1% of the values of x are greater than x0.

Answer 1

μ=50

σ=3

The Z score (Z) is obtained as:

$Z= (x- μ)/σ$

Hence

$x=Z* σ+ μ$

Solution a)

$p(X ≤ X_0 )=0.8413$

This is a left tail probability:

$p value=1-p(X ≤ X_0 )=1-0.8413=0.1587$

At $p value=0.1587$

$Z score=-1$

Therefore:

$x=(-1)* 3+ 50$

Answer: 47.00

Solution b)

$p(X> X_0 )=0.25$

This is a right tail probability:

$p value=1-p(X ≤ X_0)=1-0.25=0.75$

At $p value=0.75$

$Z score=-0.67$

Therefore:

$x=(-0.67)* 3+ 50$

Answer: 47.99

Solution c)

$p(X> X_0 )=0.95$

This is a right tail probability:

$p value=1-p(X ≤ X_0)=1-0.95=0.05$

At $p value=0.05$

$Z score=1.64$

Therefore:

$x=(1.64)* 3+ 50$

Answer:  54.92

Solution d)

$p(41 ≤X ≤ X_0 )=0.8630$

This is a two tail probability:

$p value=1-p(X ≤ X_0)=1-0.8630=0.137$

Obtain the left-side rejection region $p(41 <X)$

$Z= (x-μ)/σ= (41- 50)/3 = -3$

 

At $Z value=-3$

$p(41 <X)= 0.00135$

Hence, the probability can be expressed as a right tail probability as follows:

$p(X≤ X_0 )=(0.8630+0.00135)=0.86435$

$p value=1-p(X ≤ X_0) =1-0.86435=0.13565$

 

At $p value=0.13565$

$Z score=1.10$

Therefore:

$x=(1.10)* 3+ 50$

Answer:  54.10

Solution e)

If 10% of values of X are less than X_0, then

$p(X> X_0 )=0.90$

This is a left tail probability:

$p value=1-p(X ≤ X_0)=1-0.90=0.10$

At $p value=0.10$

$Z score=-1.28$

Therefore:

$x=(-1.28)* 3+ 50$

 

Answer = 46.16

 

Solution f)

If 1% of values of X are greater than X_0, then

$p(X< X_0 )=0.99$

This is a right tail probability:

$p value=1-p(X ≤ X_0)=1-0.99=0.01$

At $p value=0.01$

$Z score=2.33$

Therefore:

$x=(2.33)* 3+ 50$

 

Answer:  56.99