At a large dept. store, the avg. number of years of employment for a cashier is 5.7 with a standard deviation of 1.8 years. If the number of years of employment at this department store is normally distributed, what is the probability that a cashier selected at random has worked at the store for over 10 years?
Calculate the Z-score:
Z=(10-5.7)/1.8=2.39
Look up the normal distribution table for the corresponding probability for LESS than 10 years, subtract this probability from 1.0 to get the probability of service of 10 years or more.
Table link:
http://www.math.unb.ca/~knight/utility/NormTble.htm
To find the probability that a randomly selected cashier has worked at the store for over 10 years, we need to calculate the z-score and then find the corresponding probability.
The z-score formula is given by:
z = (x - μ) / σ
where:
x = 10 (number of years)
μ = 5.7 (mean)
σ = 1.8 (standard deviation)
Substituting the values, we have:
z = (10 - 5.7) / 1.8
z = 4.3 / 1.8
z ≈ 2.39
Next, we need to find the probability associated with this z-score using a standard normal distribution table or calculator.
P(Z > 2.39) ≈ 0.0081
Therefore, the probability that a randomly selected cashier has worked at the store for over 10 years is approximately 0.0081, or 0.81%.
To find the probability that a randomly selected cashier has worked at the store for over 10 years, we can use the concept of standard deviation and z-scores.
First, let's calculate the z-score for the value of 10 years using the formula:
z = (x - μ) / σ
Where:
x = The given number of years (10)
μ = The mean number of years (5.7)
σ = The standard deviation (1.8)
Using the values given, we can calculate:
z = (10 - 5.7) / 1.8
z = 4.3 / 1.8
z ≈ 2.39
Next, we can use a standard normal distribution table or a calculator that provides the cumulative probability to find the probability corresponding to the z-score of 2.39.
Using the table or calculator, we find that the area to the left of z = 2.39 is approximately 0.9918.
Since we want the probability of a value greater than 10 years, we need to subtract the cumulative probability from 1:
P(x > 10) = 1 - P(x ≤ 10)
= 1 - 0.9918
≈ 0.0082
Therefore, the probability that a randomly selected cashier at the department store has worked for over 10 years is approximately 0.0082, or 0.82%.