4. The scores on a university examination are normally distributed with a mean of 62 and a standard deviation of 11. If the bottom 5% of students will fail the examination, what is the lowest score, to the nearest whole number, that a student can have and pass?

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion. Use the corresponding Z score to find your answer.

To determine the lowest score that a student can have and still pass the examination, we need to find the cutoff score for the bottom 5% of the scores.

To do this, we'll use the concept of z-scores, which measure the number of standard deviations a particular value is from the mean in a normal distribution.

First, let's find the z-score corresponding to the bottom 5% (or the 5th percentile). This z-score will indicate the number of standard deviations below the mean that a student's score should be in order to fail the examination.

Since the distribution is normal, we know that approximately 95% of the scores are below the mean, leaving 5% of the scores above it. We want to find the z-score that corresponds to this 5th percentile.

From a standard normal distribution table or using a statistical calculator, we can find that the z-score corresponding to the 5th percentile is approximately -1.645.

Now, we can use the formula for z-score conversion to find the corresponding raw score in terms of the mean and standard deviation:

z = (x - μ) / σ

Where:
- z is the z-score
- x is the raw score
- μ is the mean
- σ is the standard deviation

Rearranging the formula to solve for x:

x = z * σ + μ

Plugging in the values, we get:

x = -1.645 * 11 + 62 ≈ 41.71

Since the lowest score needs to be rounded to the nearest whole number, the lowest passing score that a student can have is 42.

To find the lowest score that a student can have and still pass the examination, we need to determine the cutoff point for the bottom 5% of scores.

Step 1: Determine the Z-score corresponding to the bottom 5% of the distribution.

The area to the left of a z-score can be found using a standard normal distribution table or calculator. Since we are looking for the bottom 5%, we need to find the z-score that corresponds to an area of 0.05 to the left of it.

Using a standard normal distribution calculator or table, we find that the z-score corresponding to an area of 0.05 to the left is approximately -1.645.

Step 2: Use the z-score formula to find the raw score.

The formula to convert a z-score to a raw score is:
x = μ + z * σ

where x is the raw score, μ is the mean, z is the z-score, and σ is the standard deviation.

In this case, μ = 62, and σ = 11.

x = 62 + (-1.645) * 11
x ≈ 62 - 18.0955
x ≈ 43.9045

Step 3: Round the raw score to the nearest whole number.

Rounding 43.9045 to the nearest whole number gives us the lowest score that a student can have and still pass the examination, which is 44.