## Normal Distribution IB Matheamtics HL

Discussions for the Core part of the syllabus. Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths HL Revision Notes

### Normal Distribution IB Matheamtics HL

IB Mathematics HL – 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 10% of the students weigh at least 53 Kg.

Thanks
elizabeth

Posts: 0
Joined: Mon Jan 28, 2013 8:09 pm

### Re: IB Mathematics HL Normal Distribution

IB Mathematics HL – Continuous Probability Distribution, 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.
1. The mean, median, and mode are equal.
2. The normal curve is bell shaped and is symmetric about the mean.
3. The total are under the normal curve is equal to one.
4. 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.

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

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

We know that $P(W \geq 53)=0.1$

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

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

$\Rightarrow P(Z \geq \frac{53- \mu}{2})=0.1$

Using GDC Casio fx-9860G SD

Setting Tail: right
Area: 0.1
$\sigma$:1
$\mu$:0

We find that the standardized value is 1.28155157

Therefore
$\frac{53- \mu}{2}=1.28155157 \Rightarrow 53- \mu =2.56310$

$\mu=53- 2.56310=50.4 (3 s.f.)$

$\mu$

Hope these help!!
miranda

Posts: 268
Joined: Mon Jan 28, 2013 8:03 pm

### Re: Normal Distribution IB Matheamtics HL

Thank you miranda!!
elizabeth

Posts: 0
Joined: Mon Jan 28, 2013 8:09 pm