IB Math HL – Normal Distribution

A Normal distribution is a continuous probability distribution for a random variable X. The graph of a normal distribution is called the normal curve. A normal distribution has the following properties.

The mean, median, and mode are equal.
The normal curve is bell-shaped and is symmetric about the mean.
The total area under the normal curve is equal to one.
The normal curve approaches, but never touches, the x-axis as it extends farther and farther away from the mean.

Approximately 68% of the area under the normal curve is between $\mu - \sigma$ and $\mu + \sigma$ .

Approximately 95% of the area under the normal curve is between $\mu - 2 \sigma$ and $\mu +2 \sigma$ .

Approximately 99.7% of the area under the normal curve is between $\mu - 3 \sigma$ and $\mu + 3 \sigma$ .

The standard normal distribution is a normal probability distribution that has a mean of 0 and a standard deviation of 1.

Questions and Answers

1. IB Maths HL -Continuous Probability Distribution-Normal Distribution

How can we find the standard deviation of the weight of a population of dogs which is found to be normally distributed with mean 7.4 Kg and the 30% of the dogs weigh at least 8 Kg.

Solution

Let the random variable W denote the weight of the dogs, so that

$W \sim N(7.4, \sigma ^2)$

We know that $P(W \geq 8)=0.3$

Since we don’t know the standard deviation, we cannot use the inverse normal. Therefore we have to transform the random variable $W$ to that of $Z \sim N(0, 1)$ , using the transformation $Z= \frac{X- \mu}{\sigma}$ we have the following

$P(W \geq 8)=0.3 \Rightarrow P(\frac{X- \mu}{\sigma} \geq \frac{8- 7.4}{\sigma})=0.3$

$\Rightarrow P(Z \geq \frac{0.6}{\sigma})=0.3$

Using GDC Casio fx-9860G SD

MAIN MENU > STAT>DIST(F5)>NORM(F1)>InvN>

Setting Tail: right

Area: 0.3

$\sigma$ :1

$\mu$ :0

We find that the standardized value is 0.5244

Therefore, $\frac{0.6}{\sigma}=0.5244 \Rightarrow \sigma=\frac{0.6}{0.5244}=1.14 (3 s.f.)$

2. IB Mathematics HL – Statistics Probability, Normal Distribution Continuous Probability Distribution

How can we find the mean $\mu$

of the weight of a population of students which is found to be normally distributed with standard deviation 2 Kg and the 30% of the students weigh at least 53 Kg.

Solution

Let the random variable W denote the weight of the students, so that

$W\sim N( \mu, 2 ^2)$

We know that $P(W \geq 53)=0.3$

Since we don’t know the mean, we cannot use the inverse normal. Therefore we have to transform the random variable $W$ to that of

$Z\sim N(0,1)$ , using the transformation $Z= \frac{X- \mu}{\sigma}$ , we have the following

$P(W \geq 53)=0.3 \Rightarrow P(\frac{X- \mu}{\sigma} \geq \frac{53- \mu}{2})=0.3$

$\Rightarrow P(Z \geq \frac{53- \mu}{2})=0.3$

Using GDC Casio fx-9860G SD

MAIN MENU > STAT>DIST(F5)>NORM(F1)>InvN>

Setting Tail: right

Area: 0.1

$\sigma$ :1

$\mu$ :0

We find that the standardized value is 0.5244

Therefore,

$\frac{53- \mu}{2}=0.5244 \Rightarrow 53- \mu =1.0488$

$\mu=53-1.0488=51.9 (3 s.f.)$

3. IB Mathematics HL – Normal Distribution Continuous Probability Distribution

How can we find the percentage of the population will benefit from a new tax law expected to benefit families with income between $40,000 and $50,000 given that the family income follows a normal distribution with mean $45,000 and standard deviation $ 12,000?

Solution

Let the random variable I denote the family income, so that

$I\sim N( 45,000, 12,000 ^2)$

the probability is $P(40,000 \geq I \geq 50,000)$

Using GDC Casio fx-9860G SD

MAIN MENU > STAT>DIST(F5)>NORM(F1)>Ncd>

Setting Lower: 40,000

Upper: 50,000

$\sigma$ : 12,000

$\mu$ : 45,000

We find that the probability is 0.323

So 32.3% of the families will benefit from this new tax law.