(i)     Write down the probability that a fish weighs more than \(760\) g.

(ii)     Find the probability that a fish weighs less than \(815\) g.

(i)     Write down the standardized value for \(815\) g.

(ii)     Hence or otherwise, find \(\sigma \).

A fishing contest takes place in the lake. Small fish, called tiddlers, are thrown back into the lake. The maximum weight of a tiddler is \(1.5\) standard deviations below the mean.

Find the maximum weight of a tiddler.

A fish is caught at random. Find the probability that it is a tiddler.

\(25\% \) of the fish in the lake are salmon. \(10\% \) of the salmon are tiddlers. Given that a fish caught at random is a tiddler, find the probability that it is a salmon.

Note:     There may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Accept answers that are consistent with their working.

(i)     \({\text{P}}(X > 760) = 0.5{\text{ (exact), }}[0.499,{\text{ }}0.500]{\text{ }}\)     A1     N1

(ii)     evidence of valid approach     (M1)

recognising symmetry, \(\frac{{0.7887}}{2},{\text{ }}1 - {\text{P}}(W < 815),{\text{ }}\frac{{21.13}}{2} + 78.87\% \)

correct working     (A1)

eg\(\;\;\;\)\(0.5 + 0.39435,{\text{ }}1 - 0.10565,\) 

\(0.89435{\text{ (exact)}},{\text{ }}0.894{\text{ }}[0.894,{\text{ }}0.895]\)     A1     N2

[4 marks]

(i)     \(1.24999\)     A1     N1

\(z = 1.25{\text{ }}[1.24,{\text{ }}1.25]\)

(ii)     evidence of appropriate approach     (M1)

eg\(\;\;\;\)\(\sigma  = \frac{{x - \mu }}{{1.25}},{\text{ }}\frac{{815 - 760}}{\sigma }\)

correct substitution     (A1)

eg\(\;\;\;\)\(1.25 = \frac{{815 - 760}}{\sigma },{\text{ }}\frac{{815 - 760}}{{1.24999}}\)

\(44.0003\)

\(\sigma  = 44.0{\text{ }}[44.0,{\text{ }}44.1]{\text{ (g)}}\)     A1     N2

[4 marks]

correct working     (A1)

eg\(\;\;\;\)\(760 - 1.5 \times 44\)

\(693.999\)

\(694{\text{ }}[693,{\text{ }}694]{\text{ (g)}}\)     A1     N2

[2 marks]

\(0.0668056\)

\({\text{P}}(X < 694) = 0.0668{\text{ }}[0.0668,{\text{ }}0.0669]\)     A2     N2

[2 marks]

recognizing conditional probability (seen anywhere)     (M1)

eg\(\;\;\;\)\({\text{P}}({\text{A}}|{\text{B}}),{\text{ }}\frac{{0.025}}{{0.0668}}\)

appropriate approach involving conditional probability     (M1)

eg\(\;\;\;\)\({\text{P}}(S|T) = \frac{{{\text{P}}(S{\text{ and }}T)}}{{{\text{P}}(T)}}\),

correct working

eg\(\;\;\;\)P (salmon and tiddler) \( = 0.25 \times 0.1,{\text{ }}\frac{{0.25 \times 0.1}}{{0.0668}}\)     (A1)

\(0.374220\)

\(0.374{\text{ }}[0.374,{\text{ }}0.375]\)     A1     N2

[4 marks]

Total [16 marks]

There was a wide range of ability shown by candidates in this question. While the majority knew how to find probabilities, very few understood the concepts behind the normal distribution, including the answer to the straightforward question (ai). Quite a few students did not yet recognize the instruction “write down”, spending considerable time trying to find the 0.5 answer in (ai) or the standardised value in (bi).

Many candidates did not understand question (bi), giving either a probability value as the z-value or finding the correct value later on in part (bii) in the calculation of the standard deviation (without recognising its significance). For many of those who did understand these concepts, the context of the question was not a real challenge and a number of candidates managed to answer the entire question correctly.

There was a wide range of ability shown by candidates in this question. While the majority knew how to find probabilities, very few understood the concepts behind the normal distribution, including the answer to the straightforward question (ai). Quite a few students did not yet recognize the instruction “write down”, spending considerable time trying to find the 0.5 answer in (ai) or the standardised value in (bi).

Many candidates did not understand question (bi), giving either a probability value as the z-value or finding the correct value later on in part (bii) in the calculation of the standard deviation (without recognising its significance). For many of those who did understand these concepts, the context of the question was not a real challenge and a number of candidates managed to answer the entire question correctly.

There was a wide range of ability shown by candidates in this question. While the majority knew how to find probabilities, very few understood the concepts behind the normal distribution, including the answer to the straightforward question (ai). Quite a few students did not yet recognize the instruction “write down”, spending considerable time trying to find the 0.5 answer in (ai) or the standardised value in (bi).

Many candidates did not understand question (bi), giving either a probability value as the z-value or finding the correct value later on in part (bii) in the calculation of the standard deviation (without recognising its significance). For many of those who did understand these concepts, the context of the question was not a real challenge and a number of candidates managed to answer the entire question correctly.

There was a wide range of ability shown by candidates in this question. While the majority knew how to find probabilities, very few understood the concepts behind the normal distribution, including the answer to the straightforward question (ai). Quite a few students did not yet recognize the instruction “write down”, spending considerable time trying to find the 0.5 answer in (ai) or the standardised value in (bi).

Many candidates did not understand question (bi), giving either a probability value as the z-value or finding the correct value later on in part (bii) in the calculation of the standard deviation (without recognising its significance). For many of those who did understand these concepts, the context of the question was not a real challenge and a number of candidates managed to answer the entire question correctly.

There was a wide range of ability shown by candidates in this question. While the majority knew how to find probabilities, very few understood the concepts behind the normal distribution, including the answer to the straightforward question (ai). Quite a few students did not yet recognize the instruction “write down”, spending considerable time trying to find the 0.5 answer in (ai) or the standardised value in (bi).

Many candidates did not understand question (bi), giving either a probability value as the z-value or finding the correct value later on in part (bii) in the calculation of the standard deviation (without recognising its significance). For many of those who did understand these concepts, the context of the question was not a real challenge and a number of candidates managed to answer the entire question correctly.