\( \DeclareMathOperator{cosec}{cosec} \)

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Here are some exam-style questions on this statement:

- "
*The time in minutes taken to drive through a city at different times of the day can be modelled by a normal distribution with mean of 45 and standard deviation of 12. A commuter will be late for work if it takes more than an hour to drive through the city.*" ... more - "
*The travelling times by train, \( T \) minutes, between two towns can be modelled by a normal distribution with a mean of 55 minutes and a standard deviation of \( \sigma \) minutes.*" ... more - "
*The weights of players in a sports league are normally distributed with a mean of 75.2 kg, (correct to three significant figures). It is known that 75% of the players have weights between 67 kg and 80 kg. The probability that a player weighs less than 67 kg is 0.05.*" ... more - "
*The length, \(X\) minutes , of a certain category of online video is normally distributed with a mean of 28.*" ... more - "
*The random variable X follows a normal distribution with mean \(\mu\) and standard deviation \(\sigma\).*" ... more - "
*The length of Costlow's bâtard bread loaves in centimetres is normally distributed with mean \( \mu \). The following table shows probabilities for values of \(L\).*" ... more

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In statistics, standardization is the process of converting an actual data value to a z-score, which represents the number of standard deviations the data value is from the mean. This is particularly useful when comparing scores from different normal distributions. The formula for standardization is given by:

$$ z = \frac{(x - \mu)}{\sigma} $$where \( x \) is the data value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. Inverse normal calculations are used to find the data value associated with a particular cumulative probability.

The key formula for inverse normal calculations is:

$$ x = \mu + z \cdot \sigma $$For instance, consider a student's score \( x \) on a test to be 85, with the test scores normally distributed with a mean \( \mu \) of 75 and a standard deviation \( \sigma \) of 10. To standardize this score, we calculate the z-value:

$$ z = \frac{(85 - 75)}{10} \\ z = \frac{10}{10} \\ z = 1 $$This z-score of 1 indicates that the student's score is 1 standard deviation above the mean. This standardized score can now be compared across different normal distributions or used in further statistical analyses.

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