Answer to Question #213128 in Statistics and Probability for Dr. Nazir

Question

3. 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.

Answer 1

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

(a)

$P(X<15) = P(Z< \frac{15-18}{2.5}) \\ = P(Z< -1.2) \\ = 0.1151$

(b)

$P(17<X<21) = P(X<21) -P(X<17) \\ =P(Z< \frac{21-18}{2.5}) -P(Z< \frac{17-18}{2.5}) \\ = P(Z<1.2) -P(Z< -0.4) \\ = 0.8849 -0.3446 \\ = 0.5403$

(c)

$P(X<k) = 0.2578 \\ P(Z< -0.65) = 0.2578 \\ \frac{k-18}{2.5} = -0.65 \\ k = 18 - 1.65 \times 2.5 \\ k = 13.875$

(d)

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