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2 years ago

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The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives 13.113.113, point, 1 years; th

e standard deviation is 1.51.51, point, 5 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 meerkat living longer than 14.614.614, point, 6 years.

1 answer:

skad [1K]2 years ago

8 0

Answer:

The Probability = 0.16

Step-by-step explanation:

Empirical rule formula states that:

68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

Average = 13.1 years

Standard deviation = 1.5 years

x = 14.6 years

= 14.6 -13.1/1.5

= 1

This falls within 1 standard deviations of the mean

68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

Hence:

100 - 68%/2 = 32/2 = 16%

Converting to probability = 0.16

The probability of a meerkat living longer than 14.6 years is 0.16

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Answer:

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Step-by-step explanation:

Let x be the distance of a point P ( lies on the building ) from the top of the building such that AP is perpendicular to the building and y be the distance of the building from point A, ( shown in the below diagram )

Given,

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$20\text{ minutes }=\frac{20}{60}=\frac{1}{3}\text{ degree}$

$50\text{ minutes }=\frac{50}{60}=\frac{5}{6}\text{ degree}$

By the below diagram,

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$tan(12+\frac{5}{6})^{\circ}=\frac{8.20}{y}$

$tan (\frac{77}{6})^{\circ}=\frac{8.20}{y}$

$\implies y=\frac{8.20}{tan (\frac{77}{6})^{\circ}}$

Now, again by the below diagram,

$tan (31^{\circ}20')=\frac{x}{y}$

$tan(31+\frac{1}{3})=\frac{x}{y}$

$\implies x=y\times tan(\frac{94}{3})=\frac{8.20}{tan (\frac{77}{6})^{\circ}}\times tan(\frac{94}{3})^{\circ}=21.9142943216\approx 21.91$

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When 1027 is changed to 946 it will fall between 938 and 1002. So updated sorted list of numbers will now look like,

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Therefore, , median will not change.

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In the initial case,

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