Question

the value of k such that P(X<k) = 0.2578

Answer 1

Full problem:

Given the normally distributed variable X with mean 18 and standard deviation 2.5, find

(a) P(X<15):

(h) P(17< X<21);

(c) the value of k such that P(X<k) = 0.2578;

(d) the value of k such that P(X> A) = 0.1539.

$\mu=18 \\ \sigma=2.5$

(a)

$P(X<15) = P(Z< \frac{15-18}{2.5}) \\ =P(Z< -1.2) \\ = 0.1150$

(b)

$P(17<X<21) = P(X<21) -P(X<17) \\ =P(Z< \frac{21-18}{2.5}) -P(Z< \frac{17-18}{2.5}) \\ =P(Z< 0.8571) -P(Z< -0.4) \\ = 0.8042 -0.3445 \\ = 0.4597$

(c)

$P(X<k) = 0.2578 \\ P(Z< \frac{k-18}{2.5}) =0.2578 \\ \frac{k-18}{2.5} = -0.65 \\ k-18 = 2.5 \times (-0.65) \\ k=18 -1.625 \\ k=16.375$

(d)

$P(X>k) = 0.1539 \\ P(X<k) = 1 -0.1539 = 0.8461 \\ P(Z< \frac{k-18}{2.5}) =0.8461 \\ \frac{k-18}{2.5} = 1.02 \\ k-18=2.5 \times 1.02 \\ k = 18 +2.55 \\ k=20.55$