factory worker wages, according to data released by city chamber of commerce, the weekly wages of factory workers are normally distributed according to the probability density function
f(x)= (1/50√2π)e^((-1/2)((x-500)/50)^2)
find the probability that a worker selected at random from the city has a weekly wage of 450-550
From f(x) you see that 500 is the mean and 50 the standard deviation.
You can write the interval 450-550 as:
500 +/- 50 = mean +/- standard deviation
So, what's the probability a normally distributed variable to be within one standard deviation of the mean?
To find the probability that a worker has a weekly wage of 450-550, we need to integrate the given probability density function (PDF) over that range of wages.
The given PDF is: f(x) = (1/ (50√(2π))) * e^((-1/2) * ((x-500)/50)^2)
To find the probability, we integrate this function with respect to x from 450 to 550:
P(450 ≤ x ≤ 550) = ∫[450,550] f(x) dx
However, integrating this function directly might be a bit complex. To simplify, we can make use of the standard normal distribution and the properties of the z-score.
First, let's standardize the range of wages (450-550) to a standard normal distribution using the formula:
z = (x - μ) / σ
where:
x = wage (450-550)
μ = mean (500)
σ = standard deviation (50)
So, z = (450 - 500) / 50 = -1
z = (550 - 500) / 50 = 1
Now, we can use the z-table or a statistics calculator to find the cumulative probability of the standardized range (-1 ≤ z ≤ 1). This will give us the probability of having a wage between 450 and 550.
Using the z-table or calculator, we find that the cumulative probability for z = -1 is approximately 0.1587, and the cumulative probability for z = 1 is approximately 0.8413.
Therefore, the probability of a worker selected at random from the city having a weekly wage of 450-550 is:
P(450 ≤ x ≤ 550) = 0.8413 - 0.1587 = 0.6826, or 68.26%