Question

solve the problem. you may sketch a normal curve to help answer the questions.

1. on a standardized test, max scored 84, exactly one standard deviation above the mean. if the standard deviation for the test is 6, what is the mean score for the test?
2. a set of scores with a normal distribution has a mean of 50 and a standard deviation of 7. approximately what percent of scores fall in the range 36 - 64?
3. the mean of a normally distributed set of data is 52 and the standard deviation is 4. approximately 95% of all the cases will lie between which two measures?
4. a test was given to 120 students and the scores approximated a normal distribution. if the mean score was 72 with a standard deviation of 7, approximately what percent of the scores were 65 or higher?

Explanation:

Problem 5

Step1: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the raw score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Given that \(z = 1\) (since the score is one standard deviation above the mean), \(x=84\), and \(\sigma = 6\).

Step2: Substitute values into the formula

Substitute into \(z=\frac{x-\mu}{\sigma}\), we get \(1=\frac{84 - \mu}{6}\).

Step3: Solve for \(\mu\)

Multiply both sides by \(6\): \(6=84-\mu\). Then, \(\mu=84 - 6\).

Step1: Calculate the number of standard deviations from the mean

For \(x = 36\), \(z_1=\frac{36 - 50}{7}=\frac{-14}{7}=- 2\). For \(x = 64\), \(z_2=\frac{64 - 50}{7}=\frac{14}{7}=2\).

Step2: Use the empirical rule

The empirical rule (68 - 95 - 99.7 rule) states that for a normal distribution, approximately \(95\%\) of the data lies within \(z=-2\) and \(z = 2\).

Step1: Use the empirical rule

According to the empirical rule (68 - 95 - 99.7 rule), for a normal distribution, approximately \(95\%\) of the data lies within \(z=-2\) and \(z = 2\).

Step2: Calculate the values

If \(\mu = 52\) and \(\sigma=4\), then \(x_1=\mu-2\sigma=52-2\times4=52 - 8 = 44\) and \(x_2=\mu + 2\sigma=52+2\times4=52 + 8=60\)

Answer:

\(78\)

Problem 6