IB Maths SL 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 is 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 $\mu + \sigma$

and . 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.

Why is the Normal Distribution is so important?

Many things actually are normally distributed, or very close to it. For example, height and weight are approximately normally distributed.

Questions and Answers

1. IB Mathematics SL – 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 10% of the dogs weigh at least 8 Kg.

Solution

IB Mathematics SL – Continuous Probability Distribution, Normal Distribution

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

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

We know that $P(W \geq 8)=0.1$

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

$P(W \geq 8)=0.1 \Rightarrow P(\frac{X- \mu}{\sigma} \geq \frac{8- 7.4}{\sigma})=0.1$

$\Rightarrow P(Z \geq \frac{0.6}{\sigma})=0.1$

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 1.28155

Therefore, $\frac{0.6}{\sigma}=1.28155 \Rightarrow \sigma=\frac{0.6}{1.28155 }=0.468 (3 s.f.)$

2. IB Mathematics SL – Continuous Probability Distribution, Normal 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 20% of the students weigh at least 53 Kg.

Solution

IB Mathematics HL – Continuous Probability Distribution, Normal Distribution

Concerning your question

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

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

We know that $P(W \geq 53)=0.2$

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 $P(W \geq 53)=0.2 \Rightarrow P(\frac{X- \mu}{\sigma} \geq \frac{53- \mu}{2})=0.1$

$\Rightarrow P(Z \geq \frac{53- \mu}{2})=0.2$

Using GDC Casio fx-9860G SD

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

Setting Tail: right

Area: 0.2

$\sigma$ :1

$\mu$ :0

We find that the standardized value is 0.8416

Therefore,

$\frac{53- \mu}{2}=0.8416 \Rightarrow 53- \mu =1.6832$

$\mu=53- 1.6832=51.3 (3 s.f.)$

3. IB Mathematics HL – Continuous Probability Distribution, Normal Distribution

How can we find the percentage of the population will benefit from a new tax law expected to benefit families with income between $30,000 and $40,000 given that the family income follows a normal distribution with mean $36,000 and standard deviation $8,000 ??

Solution

IB Mathematics SL – Continuous Probability Distribution, Normal Distribution

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

$I\sim N( 36,000, 8,000 ^2)$

the probability is $P(30,000 \leq I \leq 40,000)$

Using GDC Casio fx-9860G SD

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

Setting Lower: 30,000

Upper: 40,000

$\sigma$ : 8,000

$\mu$ : 36,000

We find that the probability is 0.465

So 46.5% of the families will benefit from this new tax law.