Normal distribution calculations: probabilities and inverse values

The variable is $X\sim \mathrm{N}(35,6^2)$, with $\mu =35$ and $\sigma =6$.

a) We need $P(30<X<40)$, the central area. With the GDC, lower limit 30, upper limit 40, $\sigma =6$ and $\mu =35$: $$P(30<X<40)=0.595$$

b) We need the right tail $P(X>45)$. As there is no upper limit, use a very distant one, for example $35+5\cdot 6=65$: $$P(X>45)=0.0478$$

c) Left tail $P(X<28)$, with a very distant lower limit, for example $35-5\cdot 6=5$: $$P(X<28)=0.122$$

The three values are consistent with the bell: the interval from 30 to 40 is centred on the mean and contains more than half of the area, and the two tails are small.

It is $X\sim \mathrm{N}(250,4^2)$.

a) We look for $m$ with $P(X<m)=0.15$. It is an inverse calculation with the area to the left, $0.15$. With the GDC inverse normal, $\mu =250$ and $\sigma =4$: $$m=245.85\dots \approx 246\text{g}$$ It is less than the mean, as it should be, because the area to the left is less than 0.5.

b) We look for $k$ with $P(X>k)=0.02$, that is, $P(X<k)=1-0.02=0.98$. With the inverse normal: $$k=258.21\dots \approx 258\text{g}$$

c) First calculate the probability for a bag of less than 245 g, $P(X<245)=0.10565\dots$. The number $Y$ of bags of less than 245 g among 800 follows a binomial $\mathrm{B}(800,0.10565\dots )$, so $$\mathrm{E}(Y)=800\cdot 0.10565\dots =84.5$$ About 84 or 85 bags are expected; there is no need to round to a whole number.

a) We know $P(X<740)=0.04$ with $\mu =750$. To standardise, find the $z$-value that leaves an area of $0.04$ to its left, using the inverse normal of the GDC for the standard normal: $$z=-1.7507\dots$$ It is negative, as corresponds to a value below the mean. We set up $$\frac{740-750}{\sigma }=-1.7507\dots \Rightarrow \sigma =\frac{-10}{-1.7507\dots }=5.71\text{ml}$$

b) With the same $\sigma$ and the new mean ${\mu }^{\prime }$, $P(X<740)=0.01$ holds. The value of $z$ for a left area of $0.01$ is $$z=-2.3263\dots$$ Then $$\frac{740-{\mu }^{\prime }}{5.7120\dots }=-2.3263\dots \Rightarrow {\mu }^{\prime }=740+2.3263\dots \cdot 5.7120\dots =753.29\dots \approx 753\text{ml}$$ The mean is raised slightly, as expected if you want fewer bottles with a low volume.

c) The variable is now $X\sim \mathrm{N}({\mu }^{\prime },{\sigma }^2)$ with ${\mu }^{\prime }=753.29\dots$ and $\sigma =5.7120\dots$. We need the right tail $P(X>760)$. With the GDC, using lower limit 760 and a very distant upper limit: $$P(X>760)=0.120$$

In the exam: when a parameter is unknown, the marks are earned by setting up the equation with the $z$-score (for example $\frac{740-750}{\sigma }=z$) and by obtaining $z$ with the standard inverse normal; a value of $\sigma$ given without that equation can lose method marks. Check the sign of $z$, use $\sigma$ unrounded in the following parts, and give the answers to 3 significant figures with units.