The normal distribution: mean, standard deviation and standardisation

a) The bell is symmetrical about the mean $\mu =40$, so half of the area lies to the right: $$P(X>40)=\frac{1}{2}$$

b) The interval $30<X<50$ is $\mu -2\sigma <X<\mu +2\sigma$, because $40-2\cdot 5=30$ and $40+2\cdot 5=50$. By the 68-95-99.7 rule the probability should be around 0.95. The GDC gives: $$P(30<X<50)=0.954$$

c) By symmetry, the probability left outside $\mu \pm 2\sigma$ is shared equally between the two tails: $(1-0.954)/2\approx 0.023$. The GDC gives a more precise value: $$P(X>50)=0.0228$$

a) $z=\frac{88-72}{8}=\frac{16}{8}=2$. A mark of 88 is 2 standard deviations above the mean.

b) $z=\frac{60-72}{8}=\frac{-12}{8}=-\frac{3}{2}$. A mark of 60 is one and a half standard deviations below the mean.

c) Solving for $x$: $x=\mu +z\sigma =72+(-0.5)\cdot 8=68$.

d) As $z=2$, this probability coincides with $P(Z<2)$ in the standard normal. The GDC gives: $$P(X<88)=0.977$$ Interpretation: approximately 97.7% of the people who take the test score less than 88 points. This is consistent with the 68-95-99.7 rule: below $\mu +2\sigma$ lies the central 95% plus half of the remaining 5%, that is, roughly 97.5%.

e) Enter the limits 60 and 80, the mean 72 and the standard deviation 8 (not 64). The GDC gives: $$P(60<X<80)=0.775$$

a) Each mark is standardised with its own mean and standard deviation: $$z_M=\frac{68-60}{6}=\frac{4}{3}\approx 1.33z_F=\frac{75-65}{10}=1$$

b) With the GDC, entering the mean and standard deviation of each subject: $$P(M<68)=0.909P(F<75)=0.841$$ That is, she outperforms 90.9% of the group in Mathematics and 84.1% in Physics.

c) In Physics she is 10 points above the mean and in Mathematics only 8, but the Physics marks are more spread out ($\sigma =10$ compared with $\sigma =6$). If you compare in standard deviations, $z_M=\frac{4}{3}>z_F=1$: her Mathematics mark is relatively higher up. She has stood out more in Mathematics, and the percentage of students she beats (90.9% compared with 84.1%) confirms it.

d) The limits are $48=60-2\cdot 6$ and $72=60+2\cdot 6$, that is, $\mu \pm 2\sigma$. The GDC gives: $$P(48<M<72)=0.954$$ which is the precise value of the 95% rule.

In the exam: when you are asked to compare two values from different distributions, comparing their differences from the mean is not enough: you must calculate the standardised values (or the probabilities), compare those numbers and conclude in the context of the subjects. Always write down which mean and standard deviation you used in each calculation.