Question

Suppose that the temperature is normally distributed with expectation 50℃ and variance 4℃.What is the probability that the temperature T will be between 48℃ and 53℃?What is the probability that T≥52℃?

Answer 1

The probability that the temperature will be between `48^@C` and `53^@C` is `0.4649`. The probability that temperature is greater than `52^@C` is `0.3085`.

Explanation:

As the temperatures are distributed normally with expectation `50^@C` and variance `4^@C`, this means mean is `50^@C`.

To calculate desired probability, we convert this to Z-score, which is expressed in terms of standard deviation from the mean. Z-scores have a distribution with a mean of `0` and a standard deviation of `1` and for conversion `z=(x-mu)/sigma`, where `sigma` is standard deviation.

Main problem is that the standard deviation is the square root of the variance and it is expressed in the same units as the mean is, whereas the variance is expressed in squared units.

In the question variance has, however, been mentioned in same unit as mean and hence I presume questioner means standard deviation is `4^@C`.

Considering this z-score for `48^@` is `(48-50)/4=-0.50` and for `53^@`, z-score is `(53-50)/4=0.75`

In tables, z-score is given as area under normal curve between `0` for mean value and that for say `z_1`. But as it is symmetric, the probability from `z=0` and `z=+-oo` (either side) is `0.5000` each. If we know the value of z-score, tables give us area under the curve beteen that z-score and mean and sometimes between `-oo` and z-score and we can calculate probability using this as follows.

As curve is symmetric area between `z=-0.5` and `z=0.75` is sum of area between `z=0`and `z=0.5` which is `0.1915` plus area between `z=0` and `z=0.75` i.e. `0.2734` i.e. `0.1915+0.2734==0.4649`. Hence, the probability that the temperature will be between `48^@C` and `53^@C` is `0.4649`.

And for `52^@C`, z-score is `(52-50)/4=0.50`. Now as the probability that temperature is greater than mean is `0.5000` and that temperature is between `50^@C` and `52^@C` is same as between `48^@C` and `50^@C` i.e. `0.1915`.

Hence, probability that temperature is greater than `52^@C` is `0.5000-0.1915=0.3085`.