in a certain normal distribution , 1.25% of the area lies to the left of 42 , and 1.25% of the area lies to the right of 48. find population mean and sigma
I almost didn't answer your post, because "statics" relates to physics. If you want to shorten statistics, stats would be better.
To find the population mean and standard deviation (sigma), we can use the properties of the normal distribution.
Step 1: Find the z-scores corresponding to the given percentages.
The z-score represents the number of standard deviations away from the mean.
For the left side, 1.25% of the area lies to the left of 42. We need to find the z-score that corresponds to the cumulative probability of 0.0125 (1.25% of the area):
Using a standard normal distribution table or a calculator, we find that the z-score for 0.0125 is approximately -2.17.
For the right side, 1.25% of the area lies to the right of 48. We need to find the z-score that corresponds to the cumulative probability of 0.9875 (1 - 0.0125):
Using a standard normal distribution table or a calculator, we find that the z-score for 0.9875 is approximately 2.17.
Step 2: Use the z-scores to find the values of the population mean and sigma.
The z-score formula is given by z = (x - μ) / σ, where x is the observed value, μ is the population mean, and σ is the standard deviation.
Let's start with the left side:
-2.17 = (42 - μ) / σ
Next, let's consider the right side:
2.17 = (48 - μ) / σ
Step 3: Solve the equations simultaneously.
We have two equations with two unknowns (μ and σ). We can use algebraic methods to solve them.
To eliminate σ, we can divide the two equations:
-2.17 / 2.17 = (42 - μ) / σ / (48 - μ) / σ
Simplifying:
-1 = (42 - μ) / (48 - μ)
Now we can cross-multiply:
-1 * (48 - μ) = 42 - μ
Expanding and rearranging the equation:
-48 + μ = -42 + μ
The μ (population mean) cancel out, implying that the population mean is not determined by the given information.
It means that population mean can be any value.
Therefore, we cannot determine the population mean with the given information, but we know that the standard deviation (sigma) is determined by σ = (48 - μ) / 2.17.
To find the population mean and standard deviation (sigma) of a certain normal distribution, we can use the concept of Z-scores.
1. The Z-score represents the number of standard deviations an observation or data point is from the mean. It allows us to standardize and compare values from different normal distributions.
2. Let's denote the population mean as μ (mu) and the population standard deviation as σ (sigma).
3. We are given the following information:
- 1.25% of the area lies to the left of 42
- 1.25% of the area lies to the right of 48
4. By symmetry of the normal distribution, the combined area to the left of 42 and the right of 48 is 1.25% x 2 = 2.5% (0.025).
5. Now, we can find the Z-score corresponding to this combined area of 0.025. We need to find the Z-score from the standard normal distribution table or use a Z-score calculator.
Looking up the Z-score for a cumulative area of 0.025, we find it to be approximately -1.96.
6. Using the formula for Z-score:
Z = (x - μ) / σ
We can set up two equations using the given information:
- (-1.96) = (42 - μ) / σ --> Equation 1
- 1.96 = (48 - μ) / σ --> Equation 2
7. To solve for μ and σ, we need to solve the system of equations.
First, bring σ to the left hand side in Equation 1 and Equation 2:
-1.96σ = 42 - μ --> Equation 3
1.96σ = 48 - μ --> Equation 4
Then, solve for μ in terms of σ by adding Equation 3 and Equation 4:
-1.96σ + 1.96σ = (42 - μ) + (48 - μ)
0 = 90 - 2μ
2μ = 90
μ = 45
Now, substitute the value of μ into Equation 1 or Equation 2 to solve for σ:
-1.96 = (42 - 45) / σ
-1.96 = -3 / σ
σ = -3 / -1.96
σ ≈ 1.53
So, the population mean (μ) is 45, and the population standard deviation (σ) is approximately 1.53.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.0125) and its Z score.
Z = ±2.24
Mean = (48-42)/2 = ?
Z = (score-mean)/SD
Insert values for Z, score and mean to solve for SD (sigma).