The normal distribution practice questions — IB Maths AA

Write each endpoint as a whole number of standard deviations from the mean:

$$130=150-20=\mu -\sigma ,190=150+2\times 20=\mu +2\sigma$$

For a normal model, about $68\mathrm{\%}$ of the data lies within $1$ standard deviation of the mean and about $95\mathrm{\%}$ within $2$. Because the curve is symmetric about $\mu$, each of those bands splits into two equal halves at the mean:

$${\displaystyle \frac{68\mathrm{\%}}{2}}=34\mathrm{\%}\text{ from }\mu -\sigma \text{ to }\mu ,{\displaystyle \frac{95\mathrm{\%}}{2}}=47.5\mathrm{\%}\text{ from }\mu \text{ to }\mu +2\sigma$$

The interval from $130$ to $190$ is exactly these two adjacent pieces joined at the mean, so it holds $34\mathrm{\%}+47.5\mathrm{\%}=81.5\mathrm{\%}$ of the apples.

Applying the two percentages to the $800$ apples gives $34\mathrm{\%}$ of $800=272$ and $47.5\mathrm{\%}$ of $800=380$:

$$272+380=652$$

Here $n=100$ and $p=0.5$, so $$\mu =np=100\times 0.5=50,{\sigma }^2=np(1-p)=100\times 0.5\times 0.5=25.$$

Thus $\sigma =5$. Since $np=50$ and $n(1-p)=50$ are both comfortably large, $X$ is modelled by $Y\sim N(50,25)$.

The score $X$ is a whole number, so the discrete event $X\ge 60$ occupies the continuous range from $59.5$ upwards: $$P(X\ge 60)\approx P(Y>59.5)\mathrm{.}$$

Standardize that boundary: $$z=\frac{59.5-50}{5}=\mathrm{1.9.}$$

The table value is $\mathrm{\Phi }(1.9)=0.9713$, and the required region is the upper tail $1-\mathrm{\Phi }(1.9)$: $$P(X\ge 60)\approx 0.0287$$

Let $x$ be the shortest export-grade length. The longest $5\mathrm{\%}$ of stems lie above $x$, so $$P(X>x)=0.05⟹P(X<x)=0.95$$

Writing $Z={\displaystyle \frac{X-42}{6}}$, the condition becomes $\mathrm{\Phi }(z)=0.95$. The standard normal table gives the upper $5\mathrm{\%}$ critical value $z=1.645$.

Undo the standardisation: $$x=42+1.645\times 6=42+9.87=51.87$$

Correct to one decimal place: $$x\approx 51.9$$

Standardise both boundaries with $Z={\displaystyle \frac{X-250}{2}}$: $$z_1=\frac{249-250}{2}=-0.5,z_2=\frac{254-250}{2}=2$$

So $P(249<X<254)=P(-0.5<Z<2)=\mathrm{\Phi }(2)-\mathrm{\Phi }(-0.5)$.

The lower $z$ is negative, so reflect it in the symmetry of the standard normal curve: $$\mathrm{\Phi }(-0.5)=1-\mathrm{\Phi }(0.5)=1-0.6915=0.3085$$

With $\mathrm{\Phi }(2)=0.9772$ from the standard normal table, $$\mathrm{\Phi }(2)-\mathrm{\Phi }(-0.5)=0.9772-0.3085=0.6687$$

Correct to three decimal places: $$P(249<X<254)\approx 0.669$$

Here $\mu =64$, $\sigma =8$ and the observation is $x=76$.

First measure how far the observation sits from the centre of the model:

$$x-\mu =76-64=12$$

That gap of $12$ hours must now be expressed in standard deviations, so divide it by $\sigma =8$:

$$z={\displaystyle \frac{76-64}{8}}={\displaystyle \frac{12}{8}}={\displaystyle \frac{3}{2}}$$

A battery lasting $76$ hours therefore lies one and a half standard deviations above the mean of the model.

With $n=150$ and $p=0.4$, $$\mu =np=150\times 0.4=60,{\sigma }^2=np(1-p)=150\times 0.4\times 0.6=36,$$ so $\sigma =6$. Both $np=60$ and $n(1-p)=90$ are large, so $X$ is modelled by $Y\sim N(60,36)$.

The count is discrete, so the event $X\ge k$ corresponds to $Y>k-0.5$. The requirement becomes $$P(Z>\frac{k-0.5-60}{6})<\mathrm{0.05.}$$

The upper $5\mathrm{\%}$ point of the standard normal is $1.645$, and the upper tail shrinks as the boundary grows, so the condition is $$\frac{k-60.5}{6}>\mathrm{1.645.}$$

Hence $k>60.5+6\times 1.645=60.5+9.87=70.37$, and the smallest integer meeting this is $71$.

Check both candidates directly. For $k=71$: $z=\frac{70.5-60}{6}=1.75$, giving an upper tail of $1-0.9599=0.0401<0.05$. For $k=70$: $z=\frac{69.5-60}{6}=1.58$, giving $1-0.9429=0.0571$, which is not below $0.05$. $$k=71$$

Standardize the two cut-offs: $$z_{63}=\frac{63-58}{5}=1,z_{53}=\frac{53-58}{5}=-1$$ $$P(X>63)=1-\mathrm{\Phi }(1)=0.1587$$ $$P(X>53)=\mathrm{\Phi }(1)=0.8413$$ Since being large guarantees being medium, the conditional probability is $$P(\text{large}\mid \text{medium})=\frac{0.1587}{0.8413}=0.189$$