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The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives 20.520.520, point, 5 years; the standard deviation is 3.93.93, point, 9 years. Use the empirical rule (68-95-99.7\%)(68−95−99.7%)left parenthesis, 68, minus, 95, minus, 99, point, 7, percent, right parenthesis to estimate the probability of a zebra living between 16.616.616, point, 6 and 24.424.424, point, 4 years.

Answer :

fichoh

Answer:

0.67894

Step-by-step explanation:

Given that :

Lifespan of zebra is normally distributed :

Mean(μ) = 20.5

Standard deviation (σ) = 3.93

Probability of Zebra living between 16. 6 to 24.4 years

Using the empirical rule :

Z = (x - μ) / σ

For : P(Z < x ) ; X = 16.6

Z = (16.6 - 20.5) / 3.93

Z = - 0.9923

P(Z < - 0.9923) = 0.16053

For : P(Z < X) ; X = 24.4

Z = (24.4 - 20.5) / 3.93

Z = 0.9923

P(Z < 0.9923) = 0.83947

Hence,

0.83947 - 0.16053 = 0.67894

Answer:

68%

Step-by-step explanation:

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