Farmed salmon reared at a fish farm have weight, W grams, modelled by W N(1200, 200^2). (a) Find P(W>1000). [3] (b) Find the probability that a randomly chosen salmon's weight differs from the mean by more than 300 grams, that is, find P(W-1200>300). [4]

Answer: ((a)) P(W>1000)=0.8413 ((b)) P(W-1200>300)=0.1336. Setting up W N(1200,200^2), so =1200 and =200. Every probability is found via Z=W- /. Part (a): P(W > 1000) z=1000-1200/200=-200/200=-1 P(W>1000)=P(Z>-1) By symmetry of the standard normal curve, P(Z>-1)=P(Z<1)= (1). From the table, (1)=0.…

A Level Maths 9709: Probability Outside a Symmetric Interval

Setting up

$W\sim N(1200,200^2)$ , so $\mu=1200$ and $\sigma=200$ . Every probability is found via $Z=\dfrac{W-\mu}{\sigma}$ .

Part (a): P(W > 1000)

$z=\frac{1000-1200}{200}=\frac{-200}{200}=-1$

$P(W>1000)=P(Z>-1)$

By symmetry of the standard normal curve, $P(Z>-1)=P(Z<1)=\Phi(1)$ . From the table, $\Phi(1)=0.8413$ , so

$P(W>1000)=0.8413$

Part (b): P(|W − 1200| > 300)

“Differs from the mean by more than $300$ g” means $W<900$ or $W>1500$ , a symmetric two-tailed probability.

Standardise both boundaries: $z=\frac{900-1200}{200}=-1.5 \qquad \text{and} \qquad z=\frac{1500-1200}{200}=1.5$

By symmetry, each tail has the same probability: $P(W<900)=P(W>1500)=1-\Phi(1.5)=1-0.9332=0.0668$

Adding the two tails: $P(|W-1200|>300)=P(W<900)+P(W>1500)=0.0668+0.0668=0.1336$

Check: $300$ g is $1.5\sigma$ from the mean on each side, so leaving roughly $13\%$ of the probability split evenly between two moderately distant tails is a sensible order of magnitude.

Final answers

(a) $P(W>1000)=\boxed{0.8413}$
(b) $P(|W-1200|>300)=\boxed{0.1336}$