In a cooking contest, the mean score for taste was 20, with a standard deviation of 2.8. One of the contestants received a score of 14. Convert this score to a z score and tell if it is “usual” or “unusual.” (1 point)

Question

a. –2.14; unusual
b. 2.14; unusual
c. 0.56; usual
d. –0.56; usual

Answer 1

To convert the score to a z-score, you can use the formula:

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

Where:

• \(X\) is the score (14 in this case),
• \(\mu\) is the mean score (20),
• \(\sigma\) is the standard deviation (2.8).

Plugging in the values:

\[ z = \frac{(14 - 20)}{2.8} = \frac{-6}{2.8} \approx -2.14 \]

Since the z-score is approximately -2.14, we can classify it as follows:

• A z-score of less than -2 or greater than 2 is typically considered unusual in a normal distribution.

Therefore, the answer is: a. –2.14; unusual