Answer :
Final answer:
The Z-score for 66 degrees at Lake Larson is 1.6. This is found by subtracting the mean from the value and dividing by the standard deviation, indicating that 66 degrees is 1.6 standard deviations above the average temperature.
Explanation:
In this problem, you are asked to find the Z-score for a temperature of 66 degrees. The Z-score, in a normal distribution, is a measure that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. If a Z-score is 0, it indicates that the data point's score is identical to the mean score.
To calculate the Z-score, you would use the formula:
Z = (X - μ) / σ
Where:
- Z = Z-score
- X = value (in our case, the given temperature)
- μ = mean (the average temperature of Lake Larson)
- σ = standard deviation
Substituting the values into the formula would give:
Z = (66 - 58) / 5 = 1.6
So, the temperature of 66 degrees is 1.6 standard deviations above the mean temperature.
This can also be interpreted that 66 degrees falls in the area of the temperature distribution that is 1.6 standard deviations above the mean.
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