The total amount being spent on personal calls in a month by employees of a company follows a normal distribution with a mean of $900 and a standard deviation of $50. Find the probability that in a randomly selected month the amount spent on personal calls is less than $750
The standard deviation, σ, is $50. The mean is $900.
To spend less than $750 means that the spending is less than mean - 3σ.
The probability of this happening, assuming the spending follows the normal distribution curve, can be found from the normal distribution tables, for example:
http://www.math.unb.ca/~knight/utility/NormTble.htm
where the probability for the tail end of 3σ is 0.001350, or 0.135%.
To find the probability that the amount spent on personal calls is less than $750, we need to calculate the area under the normal distribution curve to the left of $750.
First, we need to standardize the value $750 using the z-score formula:
z = (x - μ) / σ
Where:
x = $750 (value we want to find the probability for)
μ = $900 (mean)
σ = $50 (standard deviation)
Substituting the values into the formula:
z = (750 - 900) / 50
= -150 / 50
= -3
Next, we need to find the cumulative probability for a z-value of -3. We can use a standard normal distribution table or a calculator to find this value.
Using a standard normal distribution table, the cumulative probability for a z-value of -3 is approximately 0.0013.
This means that the probability that in a randomly selected month the amount spent on personal calls is less than $750 is approximately 0.0013, or 0.13%.