Question

Given the normally distributed variable X with mean 18 and standard deviation 2.5, find: (a) P(X < 15); (b) 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 > k) = 0.1539

Answer 1

By condition $a = 18,\,\sigma = 2.5$ .

(a) $P(X < 15) = P( - \infty < X < 15) = \Phi \left( {\frac{{15 - a}}{\sigma }} \right) - \Phi ( - \infty ) = \Phi \left( {\frac{{15 - 18}}{{2.5}}} \right) + 0.5 = 0.5 - \Phi \left( {1.2} \right) = 0.5 - 0.3489 = {\rm{0}}{\rm{.1511}}$

Answer: $P(X < 15) = {\rm{0}}{\rm{.1511}}$

(b)

$P(17 < X < 21) = \Phi \left( {\frac{{\beta - a}}{\sigma }} \right) - \Phi \left( {\frac{{\alpha - a}}{\sigma }} \right) = \Phi \left( {\frac{{21 - 18}}{{2.5}}} \right) - \Phi \left( {\frac{{17 - 18}}{{2.5}}} \right) = \Phi \left( {1.2} \right) + \Phi (0.4) = 0.3489 + 0.1554 = {\rm{0}}{\rm{.5043}}$

Answer: $P(17 < X < 21) = 0.5043$

(c)

$P(X < k) = 0.2578 \Rightarrow P( - \infty < X < k) = 0.2578 \Rightarrow \Phi \left( {\frac{{k - 18}}{{2.5}}} \right) - \Phi ( - \infty ) = 0.2578 \Rightarrow \Phi \left( {\frac{{k - 18}}{{2.5}}} \right) + 0.5 = 0.2578 \Rightarrow \Phi \left( {\frac{{k - 18}}{{2.5}}} \right) = {\rm{ - 0}}{\rm{.2422}} \Rightarrow \frac{{k - 18}}{{2.5}} = - 0.65 \Rightarrow k - 18 = 1.625 \Rightarrow k = 19.625$

Answer: $k = 19.625$

(d)

$P(X > k) = 0.1539 \Rightarrow P(k < x < \infty ) = 0.1539 \Rightarrow \Phi \left( \infty \right) - \Phi \left( {\frac{{k - 18}}{{2.5}}} \right) = 0.1539 \Rightarrow 0.5 - \Phi \left( {\frac{{k - 18}}{{2.5}}} \right) = 0.1539 \Rightarrow \Phi \left( {\frac{{k - 18}}{{2.5}}} \right) = {\rm{0}}{\rm{.3461}} \Rightarrow \frac{{k - 18}}{{2.5}} = 1.02 \Rightarrow k - 18 = 2.55 \Rightarrow k = 20.55$

Answer: $k = 20.55$