The temperature recorded by a certain thermometer when placed in boiling water (the true temperature is 100 degrees Celsius) is normally distributed with mean (u) = 99.8 degrees Celsius and standard deviation (sigma) = 0.1 degrees Celsius. Answer the following question: What is the probability that the thermometer reading is less than 99.5 degrees Celsius?

First, we need to calculate the Z-score for 99.5 degrees Celsius. The Z-score is a standardized way of measuring how many standard deviations an element is from the mean. The formula for calculating the Z-score is:

[tex]Z = \frac{(X - \mu)}{\sigma}[/tex]

where:

[tex]X = 99.5[/tex] degrees Celsius (the value we are interested in),
[tex]\mu = 99.8[/tex] degrees Celsius (the mean of the distribution),
[tex]\sigma = 0.1[/tex] degrees Celsius (the standard deviation of the distribution).

Plug the values into the formula:

[tex]Z = \frac{(99.5 - 99.8)}{0.1}[/tex]

[tex]Z = \frac{-0.3}{0.1}[/tex]

[tex]Z = -3.0[/tex]

Now that we have the Z-score, we can use a standard normal distribution table (Z-table) to find the probability that a Z-score is less than -3.0.

From the Z-table, a Z-score of -3.0 corresponds to a probability of approximately 0.0013.

This means there is a 0.13% probability that the thermometer reading will be less than 99.5 degrees Celsius. So, the thermometer is very accurate, as readings much lower than the true temperature of 100 degrees Celsius are rare.