JC2 H2 Mathematics: Normal Distribution

Recover both mean and standard deviation

Apple masses X grams are normally distributed. Ten percent are below 80 g and fifteen percent exceed 120 g. Find the mean and standard deviation.

Inverse normal calculations use cumulative left areas. Thus the second left area is 0.85, not 0.15:

Multiply by $\sigma$ before subtracting the equations:

The mean lies between the two cutoffs and closer to 120, consistent with the unequal tail areas. Keep unrounded values for later parts.

Use symmetry before standardising

For $X\sim N(\mu,\sigma^2)$, suppose $P(X\gt k)=0.45$. Find $P(X\gt 2\mu-k)$.

The cutoffs k and $2\mu-k$ lie equal distances on opposite sides of the mean. Reflection swaps left and right tails:

A normal variable has no probability mass at a single point, so strict and non-strict endpoints give the same probability.

A combined packet mass

Packet masses are independent. Type A has distribution $N(100,9)$ g and type B has distribution $N(120,16)$ g. Find the probability that two A packets and three B packets have combined mass above 0.55 kg.

Use separate subscripts for separate packets. Convert 0.55 kg to 550 g:

By contrast, $2A+3B$ means scaling one A observation and one B observation. Its variance is $2^2(9)+3^2(16)=180$, not 66. The same random mass is being reused in a multiple.

Two visits compared with one visit

Times on two visits to shop A each have distribution $N(40,10^2)$ minutes. A visit to shop B has distribution $N(70,12^2)$ minutes. Assume all three visit times are mutually independent. Find the probability that the total at A differs from the time at B by at least 15 minutes.

The event is $|D|\geq15$, so add the two disjoint tails. Write q for this probability:

Do not double one tail: D has mean 10, so the two tails cut off by −15 and 15 are unequal. The assumption covers both A visits as well as the B visit. The variable $|D|$ itself is not normal: it cannot take negative values.

Choose enough observations for a one-sided requirement

Independent filling times have distribution $N(50,12^2)$ seconds. Find the least n such that $P(\overline X\lt 53)\geq0.90$.

This is exact for any n because the individual times are normal. The required left area is at least 0.90:

The least n is 27. At n = 26 the probability is about 0.8988; at n = 27 it is about 0.9030. The denominator for a mean is $12/\sqrt n$, not $12/n$.

The heaviest of eight components

Eight independent component masses each have distribution $N(100,10^2)$ g. Find the probability the heaviest is below 110 g, then the probability the lightest exceeds 110 g.

The heaviest is below 110 only when every mass is below 110:

Here q is the requested heaviest-mass probability. The lightest exceeds 110 only when every mass exceeds 110:

For “at least one exceeds 110”, use $1-p^8$. The maximum and minimum are not themselves normally distributed.

Count acceptable bolt–hole pairs

A bolt diameter is $N(2.65,0.03^2)$ cm and a hole diameter is $N(2.72,0.02^2)$ cm, independently. A pair is acceptable when the hole is between 0.02 cm and 0.15 cm wider than the bolt. Twenty independent pairs are packed in a box. Find the probability at least 18 are acceptable.

Now change the random variable: V counts acceptable pairs, each with the same success probability p.

There are two independence assumptions: bolt and hole diameters within a pair, and acceptance events across different pairs. Retain the unrounded normal probability when using the binomial model.

Wrapping increases every fruit’s mass by eight percent

Independent fruit masses are $A\sim N(350,196)$ g and $B\sim N(250,98)$ g. A bag contains three A fruits and two B fruits. Each wrapped fruit weighs 8% more than that fruit alone. The bag itself has negligible mass. Find the probability the bag weighs above 1.7 kg.

It is simpler to convert the event back to T than to round the variance of W:

The wrapper is tied to each fruit’s mass, so it is included by multiplying that same observation by 1.08. If the whole bag is sold at $2.90 per kg, its price is $C=0.0029W$, in dollars. Then $E(C)=0.0029E(W)$ and $\operatorname{Var}(C)=0.0029^2\operatorname{Var}(W)$. Both unit conversion and percentage change square their multiplier in a variance.