2025 NJC P2 Q11

2025 NJC P2 Q11

12 marks

The fasting glucose concentration in millimoles per litre (mmol/L) of a randomly chosen man and a randomly chosen woman is normally distributed with mean and standard deviation as given in the table below.

MeanStandard deviation
Man\(5.4\)\(0.5\)
Woman\(5.0\)\(0.3\)
  1. If one man and one woman are chosen randomly, find the probability that the man’s fasting glucose concentration exceeds the woman’s. State an assumption you make in your calculation.[3]
  2. The fasting glucose concentration of at most \(3\%\) of men exceeds \(5.4\) mmol/L by \(c\) mmol/L. Find the range of \(c\).[2]
  3. For \(k > 5.4\), explain which of the following has a larger value
    \(p_1\): the probability that the fasting glucose concentration of a randomly chosen man is greater than \(k\), or
    \(p_2\): the probability that the mean fasting glucose concentration of four randomly chosen women is greater than \(k\).[3]

Fasting glucose concentration can also be measured using milligram per decilitre (mg/dL). The formula to convert mmol/L to mg/dL is\[1 \text{ mmol/L} = 18.0 \text{ mg/dL}.\]The random variable \(G\) denotes the fasting glucose concentration, in mg/dL, of a randomly chosen man.

  1. Draw a sketch to show the distribution of \(G\), including the main features of the curve.[2]
  2. On your sketch, shade the region represented by \(P(79 < G < 106)\) and state its value.[2]

Solution:

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Answer:(a) \(0.754\), assuming independent man/woman concentrations. (b) \(c\ge\frac12z_{0.97}\), whose lower boundary is \(0.940396804\ldots\) mmol/L. (c) \(p_1>p_2\). (d) \(G\sim N(97.2,81)\); sketch shown. (e) \(0.814\).

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