Find the probability that a player weighs more than \(82{\text{ kg}}\).

(i)     Write down the standardized value, z, for \(68{\text{ kg}}\).

(ii)    Hence, find the standard deviation of weights.

To take part in a tournament, a player’s weight must be within 1.5 standard deviations of the mean.

(i)     Find the set of all possible weights of players that take part in the tournament.

(ii)    A player is selected at random. Find the probability that the player takes part in the tournament.

Of the players in the league, \(25\% \) are women. Of the women, \(70\% \) take part in the tournament.

Given that a player selected at random takes part in the tournament, find the probability that the selected player is a woman.

evidence of appropriate approach     (M1)

e.g. \(1 - 0.85\) , diagram showing values in a normal curve

\({\rm{P}}(w \ge 82) = 0.15\)     A1     N2

[2 marks]

(i) \(z = - 1.64\)     A1     N1

(ii) evidence of appropriate approach     (M1)

e.g. \( - 1.64 = \frac{{x - \mu }}{\sigma }\) , \(\frac{{68 - 76.6}}{\sigma }\)

correct substitution     A1

e.g. \( - 1.64 = \frac{{68 - 76.6}}{\sigma }\)

\(\sigma = 5.23\)     A1     N1

[4 marks]

(i) \(68.8 \le {\rm{weight}} \le 84.4\)     A1A1A1     N3

Note: Award A1 for 68.8, A1 for 84.4, A1 for giving answer as an interval.

(ii) evidence of appropriate approach     (M1)

e.g. \({\rm{P}}( - 1.5 \le z \le 1.5)\) , \({\rm{P}}(68.76 < y < 84.44)\)

\({\text{P(qualify)}} = 0.866\)     A1     N2

[5 marks]

recognizing conditional probability     (M1)

e.g. \({\rm{P}}(A|B) = \frac{{{\rm{P}}(A \cap B)}}{{{\rm{P}}(B)}}\)

\({\rm{P}}({\text{woman and qualify}}) = 0.25 \times 0.7\)     (A1)

\({\rm{P}}({\rm{woman}}|{\rm{qualify}}) = \frac{{0.25 \times 0.7}}{{0.866}}\)     A1

\({\rm{P}}({\rm{woman}}|{\rm{qualify}}) = 0.202\)     A1

[4 marks]

This question was quite accessible to those candidates in centres where this topic is given the attention that it deserves. Most candidates handled part (a) well using the basic properties of a normal distribution.

In part (b) (i), candidates often confused the z-score with the area in the table which led to a standard deviation that was less than zero in part (b) (ii). At this point, candidates “fudged” results in order to continue with the remaining parts of the question. In (b) (ii), the “hence” command was used expecting candidates to use the results of (b) (i) to find a standard deviation of 4.86. Unfortunately, many decided to use their answers and the information from part (a) resulting in quite a different standard deviation of 5.79. Recognizing the inconsistency in the question, full marks were awarded for this approach, as well as full follow-through in subsequent parts of the question.

Candidates could obtain full marks easily in part (c) with little understanding of a normal distribution but they often confused z-scores with data values, adding and subtracting 1.5 from the mean of 76.

In part (d), few recognized the conditional nature of the question and only determined the probability that a woman qualifies and takes part in the tournament.