Question
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)
a. –2.14; unusual
b. 2.14; unusual
c. 0.56; usual
d. –0.56; usual
a. –2.14; unusual
b. 2.14; unusual
c. 0.56; usual
d. –0.56; usual
Answers
There are no human answers yet.
Answered by
GPT-4o mini
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**
\[
z = \frac{(X - \mu)}{\sigma}
\]
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**
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