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In a set of average city temperatures in September with a mean of 21.02°C and a standard deviation of 2.00°C, what percentage of cities have temperatures below 18.02°C?

a) 15.87%
b) 23.85%
c) 84.13%
d) 76.15%

User Englealuze
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1 Answer

2 votes

Final answer:

To find the percentage of cities with temperatures below 18.02°C, we calculate the z-score using the provided mean and standard deviation. The z-score is -1.5. However, the correct percentage closest to the area corresponding to z=-1.5 (56.68%) is not listed in the given options, indicating a potential error in the question.

Step-by-step explanation:

To determine the percentage of cities with temperatures below 18.02°C in a dataset with a mean of 21.02°C and a standard deviation of 2.00°C, we can use the z-score formula. The z-score indicates how many standard deviations an element is from the mean. The formula for the z-score is:

z = (X - μ) / σ

Where X is the value, μ (mu) is the mean, and σ (sigma) is the standard deviation. In this case:

z = (18.02 - 21.02) / 2.00 = -1.5

Using a standard normal distribution table, a z-score of -1.5 corresponds to approximately 6.68% (or look at 50% for one side of the distribution minus 43.32%, which is the cumulative area to the right of -1.5). However, the question asks for the percentage below 18.02°C, so we look at the entire area to the left of the z-score. So we need to add 50% to 6.68%, giving us 56.68%. This suggests that answer (a) 15.87% and (b) 23.85% are incorrect, and the correct answer should be close to 56.68%, which is not among the options provided. It appears there might be an error in the question or answer choices.

User Filype
by
8.3k points
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