The temperature, \(T{ }^{\circ} \mathrm{C}\), of a freezer follows a normal distribution with mean \(-6{ }^{\circ} \mathrm{C}\) and standard deviation of \(2{ }^{\circ} \mathrm{C} .\) Calculate the probability that (a) \(T>-5\) (b) \(T<-7\) (c) \(-6

Understanding the z-score is fundamental for working with the normal distribution. The z-score represents the number of standard deviations a data point is from the mean. The formula for calculating a z-score is: \
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\( z = \frac{(X - \mu)}{\sigma} \
\
\)where \( X \) is the value of interest, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation.

To visualize it, imagine the normal distribution curve: the mean lies at the center, and the standard deviations spread out symmetrically on both sides. Calculating the z-score for a specific temperature in the freezer gives us the relative position of that temperature within the distribution. For example:

• For \( T > -5^\circ C \), the z-score is \( \frac{1}{2} \) which means the temperature is half a standard deviation above the mean.
• When analyzing \( T < -7^\circ C \), the z-score is \( -\frac{1}{2} \) indicating the temperature is half a standard deviation below the mean.
• When determining the range \( -6 < T < -3 \), we compute two z-scores: 0 for \( T = -6 \) (the mean) and \( \frac{3}{2} \) for \( T = -3 \), representing 1.5 standard deviations above the mean.

By converting temperatures to z-scores, we can then use the standard normal distribution to find probabilities.

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean (also called the expected value); a high standard deviation indicates that the values are spread out over a wider range. In the given exercise, the standard deviation of the freezer temperature is \(2^\circ C\), which tells us that the temperatures typically vary by \(2^\circ C\) from the average temperature of \(-6^\circ C\).

Standard deviation is crucial for interpreting the z-score. Since the z-score measures how many standard deviations away a value is from the mean, understanding the standard deviation allows us to comprehend the significance of a certain z-score. For instance, the closer the z-score is to zero, the closer the temperature is to the mean. Contrarily, a higher z-score in absolute terms indicates a temperature that is much higher or lower than the mean.

A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. For continuous variables like temperature, the normal distribution is one such probability distribution and is characterized by its bell-shaped curve. It is defined by two parameters: the mean and the standard deviation. The area under the curve correlates to the probability of events within a certain range.

In the context of our freezer temperature example, we're using the normal distribution to calculate the probability of the temperature being in a certain range. After calculating z-scores, we look up these scores in a z-table (or use statistical software) to find the corresponding probabilities. For example, the probability of temperature being greater than \(-5^\circ C\) (z-score of \( \frac{1}{2} \)) is found to be approximately 0.30854, which interprets that there's about a 30.854% chance that the freezer temperature will be above \(-5^\circ C\). With understanding probability distribution, one can translate z-scores into real-world probabilities of events occurring, a powerful tool in statistics.