Answer to Question #213127 in Statistics and Probability for Nazir Khan

Question

2. Given a normal distribution with = 200 and o² = 100, find

(a) the area below 214;

(b) the area above 179;

(c) the area between 188 and 206;

(d) the x value that has 80% of the area below it;

(e) the two x values containing the middle 75% of the area.

Answer 1

$\mu=200 \\ \sigma^2 = 100 \\ \sigma= 10$

(a)

$P(X<214) = P(Z< \frac{214-200}{10}) \\ =P(Z<1.4) \\ = 0.9192$

(b)

$P(X>179) = 1 -P(X<179) \\ = 1 -P(Z< \frac{179-200}{10}) \\ = 1 -P(Z< -2.1) \\ = 1 -0.01786 \\ = 0.98214$

(c)

$P(188<X<206) = P(X<206) -P(X<188) \\ =P(Z< \frac{206-200}{10}) -P(Z< \frac{188-200}{10}) \\ = P(Z< 0.6) -P(Z< -1.2) \\ = 0.72575 -0.11507 \\ = 0.61068$

(d)

$P(Z<0.8416) = 0.8 \\ \frac{x-200}{10}= 0.8416 \\ x = 200 + 10 \times 0.8416 \\ x= 208.416$

(e)

$P(-z<Z<z) = 0.75 \\ P(Z<z) -P(Z< -z) = 0.75 \\ P(Z<z) -(1 -P(Z<z)) = 0.75 \\ 2P(Z<z) -1 = 0.75 \\ 2P(Z<z) = 1.75 \\ P(Z<z) = 0.875$

The value of z that leaves an area of 0.875 to the right is z₁ = -1.15 and the left is z₂ = 1.15.

$-1.15 = \frac{x_1 -200}{10} \\ x_1 = 200 -1.15 \times 10 \\ x_1 = 188.5 \\ 1.15 = \frac{x_2 -200}{10} \\ x_2 = 200 + 1.15 \times 10 \\ x_2 = 211.5$