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Degger [83]
3 years ago
15

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.
Mathematics
1 answer:
mihalych1998 [28]3 years ago
5 0

Answer:

Step-by-step explanation:

Given that X is N(50,3)

So whenever we want score for x, we can convert this to Z and use std normal distribution table.

The formula for conversion is

Z = \frac{x-50}{3}

Using this we can find out the corresponding scores

a) P(X\leq X_0) = 0.8413

Z value = 1 and hence

X_o = 50+1(3) = 53\\

b) P(x > x0) = 0.25

z=0.5-0.0987=0.4013\\X0= 50+3(0.4013)\\= 51.2039

c) P(x > x0) = 0.95\\z=2\\X_0 = 50+6 =56

d) P(41 ≤ x < x0) = 0.8630 \\P(-3 ≤ Z

e) 10th percentile is 50-1.28(3) =46.16

f) This is 99th percentile

=50+2.33(3)\\= 56.69

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