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PLS HELP ME WITH THIS ASAP

PLS HELP ME WITH THIS ASAP-example-1

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2 votes

Answer:

0.9985

Explanation:

You want the probability that a meerkat will live less than 16.1 years if their lifespan is normally distributed with mean 10.4 years and standard deviation 1.9 years.

Z-score

The z-score of the longevity is ...

(16.1 -10.4)/1.9 = 3

The empirical rule tells us that the central 3 standard deviations from the mean enclose 99.7% of the distribution. Half the remaining distribution will lie below -3 standard deviations from the mean. This tells us that the desired probability is ...

P(Z < 3) ≈ 0.997 +0.003/2 = 0.9985

The probability a meerkat lives less than 16.1 years is estimated to be 0.9985.

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User Alexenko
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