Example: $X \sim N(30, 9)$

Find $P(X > 30)$.

Step 1: Sketch

• Sketch a normal distribution curve centered at 30.
• The probability is represented by the area under the curve to the right of 30.

Step 2: Use the calculator

• Access the calculator's distribution functions.

• Choose "Normal CD" for cumulative distribution.

• Input $\mu$ (mean), $\sigma$ (standard deviation), and bounds:

• Lower bound: 30

• Upper bound: 999999 (as an approximation of infinity)

• Input $\sigma$ (not $\sigma^2$).

Step 3: Find the result

• The probability, $P$, is calculated as 0.5.

Example: $X \sim N(20, 2^2)$

Questions

a) $P(X > 22)$

b) $P(X < 18)$

c) $P(16 \leq X \leq 18)$

(a) Find $P(X > 22)$

• Use Normal CD with Lower: 22, Upper: 999999, and Standard Deviation: 2.

• $P =$ 0.1586552539 (b) Find $P(X < 18)$:

• Use Normal CD with Lower: -999999, Upper: 18, and Standard Deviation: 2.

• $P =$ 0.1586552539 (c) Find $P(16 \leq X \leq 18)$:

• Use Normal CD with Lower: 16, Upper: 18, and Standard Deviation: 2.

• $P =$ 0.1359051219

Example: The heights of a large group of women are normally distributed with a mean of 165 cm and a standard deviation of 3.5 cm. A woman is selected at random from this group.

Questions:

a) Find the probability that she is shorter than 160 cm.

b) Steven is looking for a woman whose height is between 168 cm and 174 cm for a part in his next film. Find the proportion of women from this group who meet Steven's criteria.

c) A sample of 20 women is taken from the group. Find the probability that at least 5 of the women meet Steven's criteria.

a) Find the probability that she is shorter than 160 cm.

• Use Normal CD with Lower: -999999, Upper: 160, Mean: 165, and Standard Deviation: 3.5.

• $P =$ 0.07656372552

• (Also, we could interpret this as meaning 7.66% of women are shorter than 160 cm)

b) Steven is looking for a woman whose height is between 168 cm and 174 cm for a part in his next film. Find the proportion of women from this group who meet Steven's criteria.

• Use Normal CD with Lower: 168, Upper: 174, Mean: 165, and Standard Deviation: 3.5.

• $P =$ 0.1906189738

c) A sample of 20 women is taken from the group. Find the probability that at least 5 of the women meet Steven's criteria.

• $X \sim B(20, 0.1906)$
• $P(X \geq 5) = 1 - P(X \leq 4)$
• $P(X \leq 4) =$ 0.6703
• $P(X \geq 5) = 1 - 0.6703 =$ 0.3296