The Normal Distribution practice questions — A-Level Maths

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

With $n=150$ and $p=0.4$, $$\mu = np = 150 \times 0.4 = 60, \qquad \sigma^2 = np(1-p) = 150 \times 0.4 \times 0.6 = 36,$$ 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 $$P\left(Z > \frac{k - 0.5 - 60}{6}\right) < 0.05.$$

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

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 $0.05$. $$k = 71$$

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

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

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

$$\dfrac{68\%}{2} = 34\% \ \text{from } \mu-\sigma \text{ to } \mu, \qquad \dfrac{95\%}{2} = 47.5\% \ \text{from } \mu \text{ to } \mu+2\sigma$$

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

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

$$272 + 380 = 652$$

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

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

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

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

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

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: $$P(X \ge 60) \approx P(Y > 59.5).$$

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

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

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

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

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

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

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 $\sigma = 8$:

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

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