Standard Normal Distribution

A variable is randomly distributed $X~N(50,4^2)$. Write $P(X\ge 55)$ in terms of $\Phi \left(z\right)$ for some value of z.

Since the normal distribution is continuous, $P(X\ge 55)=P(X>55)$.

We know that the probabilities in a normal distribution add up to 1, therefore $P(X>55)=1-P(X<55)$. We then convert this distribution to a standard normal distribution.

We substitute known values into the formula $z=\frac{X-\mu }{\sigma }$.

Therefore,$z=\frac{55-50}{4}$

Hence,$P(X<55)=P\left(Z<\frac{55-50}{4}\right)=P(Z<1.25)$

Finally, $P(X\ge 55)=1-P(Z<1.25)=1-\Phi (1.25)$

This is in terms of $\Phi \left(z\right)$, therefore it answers the question.

Given that $X~N(5,5^2)$, find $P(2\le X<9)$ correct to 3 significant figures.

First, let's sketch a graph:

Graph of normal distribution with shaded areas

Next, we will convert the x-scores to z-scores.

When $x=2,z=\frac{2-5}{\sqrt{5}}=-1.342$

When $x=9,z=\frac{9-5}{\sqrt{5}}=1.789$

The area to the right of z=0 is:

$$\begin{gathered} \Phi (1.789)-\Phi \left(0\right)=\Phi (1.789)-0.5 \\ =0.463 \end{gathered}$$

The area to the left of z=0 is:

$$\begin{gathered} \Phi \left(0\right)-\Phi (-1.342)=0.5-\left(1-\Phi (1.342)\right) \\ =0.4102 \end{gathered}$$

We then add the 2 areas together

$0.463+0.4102=0.8732$

$P(2\le X<9)=0.873$, to 3 significant figures

A student scores 73% on her English exam, where the class mean was 68% and the standard deviation was 10.2%. In Biology, she scored 66%, where the class mean was 62% and the standard deviation was 6.8%.

In which subject did she perform better compared with the rest of her class?

Assume that scores for both subjects were normally distributed.

To be able to effectively compare the student's scores, they need to be standardised.

English z-score: $\frac{73-68}{10.2}\approx 0.490$

Biology z-score: $\frac{66-62}{6.8}\approx 0.588$

Interpretation – the student's z-score determines the Number of standard deviations between her score and the class mean, in both cases, above the mean.

The larger the z-score, the further above her score is from the mean, and the better she performed compared to the rest of her class.

The student performed best in Biology.

The weights of handmade necklaces in a store are known to be normally distributed with a standard deviation of 5.9g. If 15% of necklaces weigh less than 58.2g, find the mean weight of the necklaces.

Let the mean weight of the necklaces be µ. The information given can therefore be written as$X~N(\mu ,5.9^2)$

Additionally, $P(X\le 58.2)=0.15$

We now convert this x-score to a z-score.

$P\left(Z\le \frac{58.2-\mu }{5.9}\right)=0.15$

We now use the inverse normal function on the calculator, for N(0, 1²) and probability=0.15

$\frac{58.2-\mu }{5.9}\approx -1.0364$

Then, solve for µ.

$$\begin{gathered} 58.2-\mu \approx -6.1 \\ \mu \approx 64.3 \end{gathered}$$

The mean weight of the necklaces is approximately 64.3g.

The random variable $X~N(\mu ,{\sigma }^2)$. Given that $P(X<17)=0.8159$, and $P(X<25)=0.9970$, find the value of µ and σ.

To solve this question, it is necessary to convert the x-scores to z-scores and then solve the two Equations simultaneously.

$$\begin{gathered} P(X<17)=0.8159 \\ \Rightarrow P\left(Z<\frac{17-\mu }{\sigma }\right)=0.8159 \end{gathered}$$

Using the inverse normal function for both probabilities, we obtain:

$$\begin{gathered} \frac{17-\mu }{\sigma }=-0.8998 \\ \Rightarrow 17-\mu =-0.8998\sigma \end{gathered}$$ other $$\begin{gathered} \frac{25-\mu }{\sigma }=2.748 \\ \Rightarrow 25-\mu =2.748\sigma \end{gathered}$$

We then solve these two Equations simultaneously. This can be done by hand or using the simultaneous equation solver on the calculator.

We obtain: $\mu =19.0$ and $\sigma =2.20$, to 3 significant figures.