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Probability question?
Suppose the mean noon-time temperature for September days in San Diego is 24∘ and the standard deviation is 4.9. (Temperature in this problem is measured in degrees celsius)
Using Chebyshev’s theorem, what is the minimal probability (in percents) that the noon-time temperature of a september days is between 14.2∘ and 33.8∘?
On September 26, 1963, the all-time record of noon-time temperature in San Diego of 44∘ was hit. Assume the temperature distribution is symmetric around the mean, what is the Chebyshev bound for the probability of breaking (or tieing) this record?
I am having a hard time understanding this. Could someone explain to me how to do this?
1 Answer
- ?Lv 78 years agoFavorite Answer
P(X < 33.8°) = 0.97725
P(X < 14.2°) = 0.02275
P(14.2° < X < 33.8°) = 0.97725 - 0.02275 = 0.9545
P(X > 44°) = 1 - P(X < 44°) = 1 - 0.99997 = 0.00003
Just for fun
P(X < 4.4°) = P(X > 44°)
Snow in September on San Diego :)
