The 60th percentile of a normal distribution is 110. The 70th percentile is 120. Find the 90th percentile.
Use same equation and Table.
To find the 90th percentile of a normal distribution, you can use the z-score formula.
Step 1: Find the z-score corresponding to the 60th percentile.
To find the z-score, you will need to find the area under the normal distribution curve to the left of the desired percentile. In this case, the area to the left of the 60th percentile is 0.60.
Using a standard normal distribution table or a calculator, you can find the corresponding z-score for an area of 0.60, which is approximately 0.253.
Step 2: Find the raw score (x-value) corresponding to the z-score.
To find the raw score corresponding to the z-score, you will need the mean and standard deviation of the normal distribution.
However, the given information does not include the mean and standard deviation, so it is not possible to find the exact raw score.
Step 3: Find the z-score corresponding to the 70th percentile.
Following the same process as in Step 1, you can find the z-score for the 70th percentile which corresponds to an area of 0.70. The z-score is approximately 0.524.
Step 4: Find the raw score (x-value) corresponding to the z-score.
Again, without the mean and standard deviation, it is not possible to find the exact raw score.
Step 5: Use the z-scores found in steps 1 and 3 to find the z-score for the 90th percentile.
Since the standard normal distribution is symmetric, you can use the fact that the difference between the z-scores for the 70th and 60th percentiles will be the same as the difference between the z-score for the 90th percentile and the z-score for the 70th percentile.
In this case, the difference between the z-scores is 0.524 - 0.253 = 0.271.
Therefore, the z-score for the 90th percentile is the z-score for the 70th percentile plus 0.271, which is approximately 0.524 + 0.271 = 0.795.
Step 6: Find the raw score (x-value) corresponding to the z-score.
Again, without the mean and standard deviation, it is not possible to find the exact raw score.
Conclusion:
Without the mean and standard deviation, it is not possible to find the exact raw score corresponding to the 90th percentile.
To find the 90th percentile of a normal distribution, we can use a standard score, also known as a z-score. The z-score indicates how many standard deviations above or below the mean a particular value is.
First, we need to find the mean and standard deviation of the normal distribution. Given that the 60th percentile is 110 and the 70th percentile is 120, we can work backwards using the z-score formula to find the mean (μ) and standard deviation (σ).
Let's assume that the 60th percentile corresponds to a z-score of 0.25 and the 70th percentile corresponds to a z-score of 0.52.
Using a z-table, we can find the values of μ and σ that lead to z-scores of 0.25 and 0.52:
For a z-score of 0.25, the z-table tells us that the corresponding area under the curve is 0.5987. This means that 59.87% of the data lies below this value. Since the 60th percentile corresponds to 59.87%, we can assume that the mean is close to 110.
Similarly, for a z-score of 0.52, the z-table gives us an area of 0.6985, which corresponds to the 70th percentile. Using this information, we can infer that the mean is close to 120.
Now that we have estimates for the mean, we can use the formula z = (x - μ) / σ, where x is the value we want to find the percentile for, to find the standard deviation.
Plug in the values:
0.52 = (120 - 110) / σ
Rearranging the equation, we get:
0.52σ = 10
Solving for σ, we obtain:
σ = 10 / 0.52 ≈ 19.23
Now that we have estimated the standard deviation (σ ≈ 19.23) and the mean (μ ≈ 110), we can find the z-score corresponding to the 90th percentile.
Using the formula again:
z = (x - μ) / σ
Plugging in the known values:
z = (x - 110) / 19.23
Rearranging the equation to solve for x:
x = z * σ + μ
Now we need to find the z-score that corresponds to the 90th percentile. This z-score can be found by referring to the z-table or using a calculator with normal distribution functionality.
Looking up the z-score for the 90th percentile, we find it is approximately 1.28. Plugging in this value, along with the mean and standard deviation, into the equation, we get:
x = 1.28 * 19.23 + 110 ≈ 134.22
Therefore, the 90th percentile of this normal distribution is approximately 134.22.