Questions: Mr. Emmer gave a test in his Chemistry class. The scores were normally distributed with a mean of 82 and a standard deviation of 4. A student is randomly chosen. What is the probability that the student scores a 70 or below? Use the formula for a z-score z=(x-μ)/σ where x is the given value, μ is the mean and σ is the standard deviation. Then refer to the chart on page 11 of the lesson to find the probability. a. 0013 b. 0179 c. 0668 d. 5000

Solution Steps

To find the probability that a student scores 70 or below, we first calculate the Z-score using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \( X = 70 \) (the score we are interested in),
• \( \mu = 82 \) (the mean score),
• \( \sigma = 4 \) (the standard deviation).

Substituting the values, we have:

\[ z = \frac{70 - 82}{4} = \frac{-12}{4} = -3.0 \]

Thus, the Z-score for the value 70 is \( z = -3.0 \).

Next, we calculate the probability that a student scores 70 or below. This is represented as:

\[ P(X \leq 70) = \Phi(z_{end}) - \Phi(z_{start}) \]

In this case, \( z_{end} = -3.0 \) and \( z_{start} = -\infty \). The cumulative distribution function \( \Phi(z) \) gives us the probability associated with a Z-score. Therefore, we have:

\[ P(X \leq 70) = \Phi(-3.0) - \Phi(-\infty) \]

From standard normal distribution tables or calculations, we find:

\[ \Phi(-3.0) \approx 0.0013 \quad \text{and} \quad \Phi(-\infty) = 0 \]

Thus, the probability is:

\[ P(X \leq 70) = 0.0013 - 0 = 0.0013 \]

Final Answer