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The random variable Y has a normal distribution with mean \( \mu \) and standard deviation \( \sigma \)\nThe \( P(Y > 17) = 0.4 \)\nFind\n(a) \( P(\mu < Y < 17) \)\n(b) \( P(\mu - \sigma < Y < 17) \) - Edexcel - A-Level Maths: Statistics - Question 3 - 2018 - Paper 1
Question 3
The random variable Y has a normal distribution with mean \( \mu \) and standard deviation \( \sigma \)\nThe \( P(Y > 17) = 0.4 \)\nFind\n(a) \( P(\mu < Y < 17) \)\n... show full transcript
Worked Solution & Example Answer:The random variable Y has a normal distribution with mean \( \mu \) and standard deviation \( \sigma \)\nThe \( P(Y > 17) = 0.4 \)\nFind\n(a) \( P(\mu < Y < 17) \)\n(b) \( P(\mu - \sigma < Y < 17) \) - Edexcel - A-Level Maths: Statistics - Question 3 - 2018 - Paper 1
Step 1
(a) P(μ < Y < 17)
Answer
To find ( P(\mu < Y < 17) ), we use the properties of the normal distribution. Given that ( P(Y > 17) = 0.4 ), we can express this as ( P(Y < 17) = 1 - P(Y > 17) = 1 - 0.4 = 0.6 ).
The probability can be standardized to standard normal form:\n[ P(\mu < Y < 17) = P(Y < 17) - P(Y < \mu) ] Which leads to:\n[ = 0.6 - 0.4 = 0.1 ]
Step 2
(b) P(μ - σ < Y < 17)
Answer
For the second part, starting with the normalization based on the standard normal variable ( Z = \frac{Y - \mu}{\sigma} ):\n[ P(\mu - \sigma < Y < 17) = P\left(\frac{\mu - \sigma - \mu}{\sigma} < Z < \frac{17 - \mu}{\sigma}\right)]\nThis becomes:\n[ P(-1 < Z < \frac{17 - \mu}{\sigma}) ]
From the first part, we know that:\n[ P(Y < \mu) = 0.5 \Rightarrow P(Y < 17) = 0.6 ]
Using the standard normal table for ( Z < 0.841 ):\nTo find it explicitly, we know that:\n[ P(Y < 17) = 0.6 ] So here:\n[ = P(-1 < Z < 0.841) = 0.841 - 0.158 = 0.683 ]
Thus, the final answer is approximately ( 0.441 ).