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The lifetime, X days, of a particular insect is such that \(\log_{10} X\) has a normal distribution with mean 1.5 and standard deviation 0.2 - CIE - A-Level Further Maths - Question 6 - 2010 - Paper 1

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The lifetime, X days, of a particular insect is such that \(\log_{10} X\) has a normal distribution with mean 1.5 and standard deviation 0.2. Find the median lifetim... show full transcript

Worked Solution & Example Answer:The lifetime, X days, of a particular insect is such that \(\log_{10} X\) has a normal distribution with mean 1.5 and standard deviation 0.2 - CIE - A-Level Further Maths - Question 6 - 2010 - Paper 1

Step 1

Find the relation for median M:

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Answer

Given that the logarithm of the lifetime ( (\log_{10} X)) follows a normal distribution, the median of (X) will be (M = 10^{\mu}), where (\mu) is the mean of the distribution. Here, (\mu = 1.5), hence:

M=101.531.62M = 10^{1.5} \approx 31.62

Thus, the median lifetime is approximately 31.62 days.

Step 2

Find also P(X > 50):

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Answer

To find (P(X > 50)), we need to convert this into the corresponding logarithmic expression:

  1. Calculate (\log_{10}(50)): log10(50)1.699\log_{10}(50) \approx 1.699

  2. We can then standardize this value to use the properties of the normal distribution: (Z = \frac{\log_{10}(50) - \mu}{\sigma} = \frac{1.699 - 1.5}{0.2} = 0.995)

  3. Now, we use the cumulative distribution function (CDF) for the standard normal distribution: P(X>50)=1Φ(0.995)10.83990.1601P(X > 50) = 1 - \Phi(0.995) \approx 1 - 0.8399 \approx 0.1601

Thus, (P(X > 50) \approx 0.16).

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