The next three questions refer to the following situation. Data from previous years reveal that the distribution if first exam scores in an introductory statistics class is approximately normal with a mean of 72 and a standard deviation of 12.

16) Given that the passing score is 65, approximately what percentage of students passes the first exam?

17) What is the probability that the student will score an “A” on the exam? (A score of 90 or more)

18) The middle 95% of the scores will tend to fall between _________ and ________.

16. Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to to the Z score.

17. Use same equation and table.

18. Use table to find P = ± ,45 to get the two Z scores. Insert into equation above.

To answer these questions, we need to use the properties of the normal distribution. Here's how you can find the answers:

16) Given that the passing score is 65, we need to find the percentage of students with a score above 65. We can do this by calculating the area under the normal curve to the right of 65.

To find this, we can use a standard normal distribution table, also known as a Z-table. First, we need to convert the passing score of 65 into a standardized z-score using the formula:

Z = (X - mean) / standard deviation

Z = (65 - 72) / 12 = -0.58

Now, we can look up the area to the right of -0.58 in the Z-table. The corresponding area is 0.7190.

To find the percentage of students who passed, we need to subtract this area from 1 and then multiply by 100:

Percentage of students passing = (1 - 0.7190) * 100 = 28.1%

Therefore, approximately 28.1% of students passed the first exam.

17) To find the probability of scoring an "A" (score of 90 or more), we need to calculate the area under the normal curve to the right of 90.

First, we need to convert the score of 90 into a standardized z-score:

Z = (90 - 72) / 12 = 1.5

Now, we can look up the area to the right of 1.5 in the Z-table. The corresponding area is 0.0668.

Therefore, the probability of scoring an "A" on the exam is approximately 0.0668, or 6.68%.

18) The middle 95% of the scores refers to the range in which 95% of the scores fall. In a normal distribution, this is also known as the 95% confidence interval.

To find the values for this interval, we can use the properties of the standard normal distribution. The middle 95% of the scores will fall between the z-scores that correspond to the cumulative probability of 0.025 and 0.975.

Using the Z-table, we find the z-scores for these cumulative probabilities to be -1.96 and 1.96, respectively.

Now, we can convert these z-scores back into scores using the formula:

X = mean + (Z * standard deviation)

Lower limit score: 72 + (-1.96 * 12) = 48.32
Upper limit score: 72 + (1.96 * 12) = 95.68

Therefore, the middle 95% of the scores will tend to fall between approximately 48.32 and 95.68.