Example: $X \sim N(25, 4)$

a) Find a guess that $P(X < a) = 0.27$

Step 1: If not directly given, work out the area to the left of the unknown boundary and sketch.

• Sketch: A normal distribution curve with mean at $25$.
• The area to the left of point a (denoted by the curve) is $0.27$.

Step 2: Input all of this information into the inverse normal function on the calculator. [Remember: "Area" means area to the left]

• Inverse Normal on Calculator:
• Area: $0.27$
• $\sigma$: $2$
• $\mu$: $25$
• $xinv$ = $23.77437427$
• So, $a = 23.77$

b) $P(X > b) = 0.42$

• [Remember: calculator only deals with the area to the left]
• Sketch: A normal distribution curve with mean at $25$. The area to the right of point $b$ is $0.42$, so the area to the left is $1 - 0.42 = 0.58$.
• Inverse Normal on Calculator:
• Area: $0.58$
• $\sigma$: $2$
• $\mu$: $25$
• $xinv = 25.40378694$

(c) $P(24 < X < c) = 0.62$

• [Need to know all areas to the left of the boundary $c$, so must calculate this area to find this.]
• $P(X \leq 24)$ = (Calculator shows 0.3085375383)
• Therefore, $P(X \leq c) = 0.62 + 0.3085 = 0.9285$
• [Put this number in inverse function]
• Inverse Normal on Calculator:
• Area: $0.9285$
• $\sigma$: $2$
• $\mu$: $25$
• $xinv = c = 27.92942048$

Example: The masses, $Y$ grams, of a brand of chocolate bar are modelled as $Y \sim N(60, 2^2)$.

a) Find the value of y such that $P(Y > y) = 0.2$

Inverse Normal on Calculator:

• Area: $0.8$
• $\sigma$: $2$
• $\mu$: $60$
• $xinv = y = 61.68324169$

b) Find the $10$% to $90$% interpercentile range of masses.

$10^{th}$ percentile:

Inverse Normal on Calculator:

• Area: $0.1$
• $\sigma$: $2$
• $\mu$: $60$
• $xinv = 57.43689672$

c) Tom says that the median is equal to the mean. State, with a reason, whether Tom is correct.

$90^{th}$ percentile:

Inverse Normal on Calculator:

• Area: $0.9$
• $\sigma$: $2$
• $\mu$: $60$
• $xinv = 62.56310328$ IPR: 62.563 - 57.637 = 5.126

Tom is correct as the normal distribution is symmetrical about the mean.

Examples:

$P(X > 38) \quad P(Y \geq 26) \quad P(Z \geq 2)$

• Graphical Representations: All show an area to the right of the value.

• Probability: All equal $0.0228.$ Explanation:

• The reason these answers are all the same is that their boundaries are exactly $2$ standard deviations from the mean.

• The number of standard deviations from the mean in a Normal Distribution is known as the z-value.

Example: Calculate the z-value for $X \sim N(27, 40)$ where $X = 18$.

Use this formula:

Substitute in the values and calculate:

Example: Find $\mu$ for $X \sim N(\mu, 12)$ given that $P(X > 6) = 0.72$.

Step 1: Calculate z-value for given boundary algebraically:

Step 2: Calculate the corresponding z-value for $Z \sim N(0,1)$ using given probability

Given:

• Graphical representation shows the area under the curve up to z.
• Using Inverse Normal calculation: $z = -0.5828414022$

Step 3: Solve the two equations for z simultaneously to find the unknown.

Calculation:

Example: For: $X \sim N(20, \sigma^2),\ given\ that\ P(X > 22) = 0.3$, find the standard deviation $\sigma$.

Step 1: Calculate z-value

Step 2: Graphical representation shows the area of $0.3$ to the right of the z-value.

Step 3: Using Inverse Normal calculation

$z = 0.5244$

Step 4: Solving for $\sigma$

Q4. (OCR 4733, Jan 2008, Q1)

The random variable T is normally distributed with mean $\mu$ and standard deviation $\sigma$. It is given that $P(T > 80) = 0.05$ and $P(T > 50) = 0.75$.

Question: Find the values of $\mu$ and $\sigma$.

Given: $T \sim N(\mu, \sigma^2)$

For $P(T > 80) = 0.05$:

The graphical representation shows the area to the right of $80$, which is $0.05$.

Using the Inverse Normal function:

$z = 1.644853667$

The equation:

Solves to:

For $P(T > 50) = 0.75$:

The graphical representation shows the area to the right of $50$, which is $0.75.$

Using the Inverse Normal function:

$z = -0.674489579$

The equation:

Solves to:

These two equations can be solved using calculator functions to find: