The length of Paulo’s lunch break follows a normal distribution with mean μ minutes and standard
deviation 5 minutes. On one day in four, on average, his lunch break lasts for more than 52 minutes.
Find the value of μ.
Z(.674) = 0.25
So, (52-μ)/5 = .674
μ = 52-5*0.674 = 48.63
To find the value of μ, we need to use the properties of a normal distribution and the information given.
Let's assume X is the length of Paulo's lunch break.
Given:
- The mean of X, denoted as μ, is unknown.
- The standard deviation of X is 5 minutes.
We are told that on one day in four, on average, his lunch break lasts for more than 52 minutes.
This information can be translated into a percentile or probability. We can use the z-score formula and the standard normal distribution to find this probability.
The z-score formula is given by:
z = (X - μ) / σ
where X is the given value (52 minutes), μ is the mean, and σ is the standard deviation.
We can rewrite the formula as:
(52 - μ) / 5 = z
To find the percentile or probability associated with a z-score, we can use a standard normal distribution table or a calculator. The z-score that corresponds to a probability of 1/4 (one day in four) is approximately -0.674.
Plugging this value into the z-score formula, we get:
(52 - μ) / 5 = -0.674
Now, we can solve for μ:
52 - μ = -0.674 * 5
52 - μ = -3.37
μ = 52 + 3.37
μ ≈ 55.37
So, the value of μ is approximately 55.37 minutes.
To find the value of μ, we need to use the properties of a normal distribution and the given information about Paulo's lunch break.
Let's break down the problem step by step:
Step 1: Set up the problem
We know that Paulo's lunch break follows a normal distribution with mean μ and standard deviation 5 minutes. Let's assume this distribution is denoted by N(μ, 5).
Step 2: Calculate the z-score
To determine the probability, we need to find the corresponding z-score. We can use the standard normal distribution table or a calculator for this.
The formula for calculating the z-score is: z = (x - μ) / σ
Where:
x = value of interest (52 minutes in this case)
μ = mean
σ = standard deviation
In this case, z = (52 - μ) / 5
Step 3: Calculate the probability
Now, we are given that on one day in four, on average, Paulo's lunch break lasts for more than 52 minutes. This translates to a probability of 1/4, or 0.25.
Since we're dealing with a normal distribution, we need to find the z-score that corresponds to a cumulative probability of 0.25 using the standard normal distribution table or a calculator.
Step 4: Find μ
Now, we need to solve for μ in the equation z = (52 - μ) / 5, where z is the z-score corresponding to a cumulative probability of 0.25.
Plug in the value of z we found in the previous step and solve for μ:
0.25 = (52 - μ) / 5
Multiply both sides by 5:
1.25 = 52 - μ
Rearrange the equation to solve for μ:
μ = 52 - 1.25
μ = 50.75
Therefore, the value of μ is approximately 50.75 minutes.