I'm kinda stuck on my math homework, does anyone know how to do it?
The results of a blood test at a medical lab are normally distributed with a mean of 60 and a standard deviation of 15. How low must a score be to lie in the lowest 5% of the results?
Look up a normal distribution table for the z-score for 5%. It is approximately -1.65 (you'll have to look it up and refine this value).
This means that all results lower than X=mean - 1.65*σ
= 60-1.65*15
= 35.25 approximately.
will lie within the lowest 5%, if the results are normally distributed.
To find the score that lies in the lowest 5% of the results, you will need to calculate the z-score corresponding to the 5th percentile. Here are the steps to follow:
Step 1: Identify the given information:
Mean (μ) = 60
Standard Deviation (σ) = 15
Percentile = 5%
Step 2: Find the z-score using a z-table or a calculator:
A z-score represents the number of standard deviations a value is from the mean. The formula to calculate the z-score is:
z = (score - mean) / standard deviation
In this case, since we want to find the score that corresponds to the 5th percentile (which is on the left side of the distribution), we need to find the z-score that corresponds to an area of 0.05.
Using a z-table, you can find the z-score corresponding to the 5th percentile as approximately -1.645.
Step 3: Calculate the score:
Now that you have the z-score, you can use it to find the corresponding raw score using the formula:
score = z-score * standard deviation + mean
Substituting the values into the formula:
score = -1.645 * 15 + 60
Calculate:
score = -24.675 + 60
Final Step:
score ≈ 35.325
Therefore, a score must be approximately less than or equal to 35.325 to lie in the lowest 5% of the results.
To find the score that lies in the lowest 5% of the results, we can use the concept of z-scores and the standard normal distribution.
First, let's understand what a z-score is. A z-score, or standard score, measures how many standard deviations a particular value is away from the mean of a distribution. It helps us compare values from different normal distributions.
To calculate the z-score for this problem, we use the formula:
z = (x - μ) / σ
where:
- x is the value we want to find the probability for (in this case, the score),
- μ is the mean of the distribution, and
- σ is the standard deviation of the distribution.
In this case, we want to find the score that lies in the lowest 5%. That means we are looking for the z-score that corresponds to a cumulative probability (area under the curve) of 5%.
To find this z-score, we can use a standard normal distribution table or a statistical calculator. Since using a table might be more time-consuming, I'll explain how to calculate it using a statistical calculator such as Excel or an online z-score calculator.
1. Go to an online z-score calculator or use a statistical software such as Excel.
2. Enter the mean (μ = 60) and the standard deviation (σ = 15).
3. Find the cumulative probability (P) that corresponds to the lowest 5% (P = 0.05).
4. Calculate the z-score (z) using the inverse of the cumulative distribution function (CDF) or the quantile function. The formula is z = invNorm(P, μ, σ).
By entering the values into the calculator or using Excel, you will find that the z-score is approximately -1.645.
Now that we have the z-score, we can solve for the score (x) using the z-score formula. Rearranging the formula, we get:
x = z * σ + μ
Plugging in the values, we have:
x = -1.645 * 15 + 60
Simplifying the equation gives us:
x ≈ 34.82
Therefore, a score must be approximately 34.82 or lower to lie in the lowest 5% of the blood test results.