14D - Finding the mean and standard deviation
When given information relating to normal distribution it is possible to determine the values of \(\mu\) and \(\sigma\).
- For each unknown, one piece of information must be given to develop an equation to solve.
- The CAS command invNormCDf is often used to find corresponding \(z\)-scores.
- The formula for standardising can then be used to determine the value of \(\mu\) and/or \(\sigma\): \[z=\frac{x-\mu}{\sigma}\]
14D - Example 1: Finding the standard deviation for a normal distribution
A normal random variable has a mean of 42 and an unknown standard deviation. Given that \(Pr(X>34)=0.812\), find the standard deviation correct to 2 decimal places.
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14D - Example 1: Video solution
Video solution coming soon! \[\sigma=9.04\] 14D - Example 1: Practice
Question 1: ABC 14D - Example 1: Solutions
Question 1: ABC |
14D - Example 2: Finding the mean and standard deviation for a normal distribution
If \(X\) is a normal random variable with \(Pr(X<31)=0.8667\) and \(Pr(X>19.5)=0.9257\), determine the mean and standard deviation. State your answers correct to 2 decimal places.
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14D - Example 2: Video solution
Video solution coming soon! \[\mu=26.00 , \sigma=4.50\] 14D - Example 2: Practice
Question 1: ABC 14D - Example 2: Solutions
Question 1: ABC |
14D - Example 3: Finding the mean for a normal distribution (VCAA, 2002)
[VCAA, 2002 Exam 1 Part 1 Question 26]
Black Mountain coffee is sold in packets labelled 250 grams. The packing process produces packets whose weight is a normally distributed random variable with a standard deviation of 3 grams. In order to guarantee that only 1% of packages are under the labelled weight, the actual mean weight (in grams) would be required to be closest to A. \(243\) B. \(247\) C. \(250\) D. \(254\) E. \(257\) |
14D - Example 3: Video solution
Video solution coming soon! Option E is correct. 14D - Example 3: Practice
Question 1: ABC 14D - Example 3: Solutions
Question 1: ABC |
14D - Example 4: Finding the mean and standard deviation for a normal distribution (VCAA, 2005)
[VCAA, 2005 Exam 1 Part 2 Question 1]
The distribution of scores in a particular examination follows a normal distribution. 30% of scores are less than 45, and 30% of scores are greater than 55. Find the mean and standard deviation, correct to one decimal place, of the scores in the examination. |
14D - Example 4: Video solution
Video solution coming soon! \[\mu=50.0 , \sigma=9.5\] 14D - Example 4: Practice
Question 1: ABC 14D - Example 4: Solutions
Question 1: ABC |
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