Standardisation and finding μ and σ

Here $\mu =60$ and $\sigma =8$.

a) $z=\frac{72-60}{8}=1.5$: the score is one and a half standard deviations above the mean.

b) $P(X<72)=P(Z<1.5)=0.933$ (normal cumulative distribution, with either scale).

c) $x=\mu +z\sigma =60+(-0.5)\cdot 8=56$.

Answer: a) $z=1.5$; b) $0.933$; c) $56$.

a) $$z_{\text{Ana}}=\frac{78-65}{10}=1.3z_{\text{Beto}}=\frac{71-60}{8}=1.375$$

b) $1.375>1.3$: although Ana has the better raw mark, Beto is further from the mean of his group, measured in standard deviations. Beto stands out more.

c) $P(\text{Physics}>71)=P(Z>1.375)=0.0846$, that is, 8.46% of students.

Answer: a) $z_{\text{Ana}}=1.3$, $z_{\text{Beto}}=\frac{11}{8}$; b) Beto; c) $0.0846$.

a) We have $P(Y<20)=0.1$. With the standard inverse normal, the area $0.1$ corresponds to $z=-1.2816$. Thus, $$\frac{20-25}{\sigma }=-1.2816\Rightarrow \sigma =\frac{5}{1.2816}=3.90$$

b) We have $P(X<40)=0.2$ and $P(X>58)=0.1$, that is, $P(X<58)=0.9$. The $z$-values for these areas are $$z_1=-0.8416(\text{area }0.2)z_2=1.2816(\text{area }0.9)$$ The two equations are set up: $$\{\begin{array}{rl}\frac{40-\mu }{\sigma }& =-0.8416\\ \frac{58-\mu }{\sigma }& =1.2816\end{array}⟹\{\begin{array}{rl}40-\mu & =-0.8416\sigma \\ 58-\mu & =1.2816\sigma \end{array}$$ Subtracting the first from the second: $18=2.1232\sigma$, so $\sigma =8.48$. Substituting, $\mu =40+0.8416\cdot 8.478=47.1$.

c) With $X\sim N(47.14;8.478^2)$: $P(X>50)=0.368$.

Answer: a) $\sigma =3.90$; b) $\mu =47.1$ and $\sigma =8.48$; c) $0.368$.

In the exam: write each $P(X<x)=p$ and its corresponding $z$ before setting up the system. Do not round $\mu$ or $\sigma$ before using them in part c): keep their full values in the calculator.