In a distribution of scores with a mean of 234 and a standard deviation of 47, what is in the 43rd percentile ? and what formula do i use ?

225.209 :]

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion and its Z score. Insert the values in the equation below and solve for the score.

Z = (score-mean)/SD

To find the 43rd percentile in a distribution, you can use the following formula:

Percentile = (P/100) * (N + 1),

Where:
- Percentile is the desired percentile (in this case, 43rd percentile)
- P is the percentile rank (43)
- N is the total number of observations in the distribution

In your case, since you only have the mean and standard deviation, you do not have the complete dataset. However, assuming that the distribution follows a normal distribution, you can use the properties of the normal distribution to approximate the percentile.

For a normal distribution, approximately 68% of the data falls within one standard deviation from the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

To find the value corresponding to the 43rd percentile, you can use the Z-score formula:

Z-score = (X - μ) / σ,

Where:
- X is the value you want to find
- μ is the mean (234)
- σ is the standard deviation (47)

First, find the Z-score corresponding to the 43rd percentile using a Z-table or a Z-score calculator. For the 43rd percentile, the Z-score is approximately -0.182.

Next, rearrange the Z-score formula to solve for X:

X = (Z * σ) + μ

Substitute the values:

X = (-0.182 * 47) + 234

Calculating this expression gives:

X ≈ 225.856

Therefore, in this distribution, the value at the 43rd percentile is approximately 225.856.