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3 years ago

Complete parts (a) and (b) below.

Mathematics

1 answer:

jeka943 years ago

7 0

Answer:

a) The mean of the probability distribution = 0.6

b) The variance of the probability distribution = 0.3

c) The standard deviation of the probability distribution = 0.5

Step-by-step explanation:

a) Data and Calculations:

number of dogs per household in a small town

Dogs Probability Expected Mean Diff Mean Diff.

Value Squared

0 0.650 0 -0.569 0.32376

1 0.216 0.216 -0.353 0.12461

2 0.087 0.174 -0.395 0.15603

3 0.026 0.078 -0.491 0.24108

4 0.014 0.056 -0.513 0.26317

5 0.009 0.045 -0.524 0.27458

The mean = 0.569 (Sum of expected value)

Variance = mean of squared differences = 0.27665 (1.38323/5)

Standard deviation = square root of variance = 0.526 = 0.5

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3 years ago

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Step-by-step explanation:

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3 years ago

A fruit company has 20% returns in periods of normal rainfall and -3% returns in droughts. The probability of normal rainfall is

WARRIOR [948]

Answer:

$E(x) = 10.8\% \\\\$

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Step-by-step explanation:

The expected returns of the fruit company is given by

$E(x) =\sum (x \cdot P(x)) \\\\$

For the given case,

Returns in normal rainfall = x₁ = 20% = 0.20

Returns in drought = x₂ = -3% = -0.03

Probability of normal rainfall = P(x₁) = 60% = 0.60

Probability of drought = P(x₂) = 40% = 0.40

So, the expected value of returns is

$E(x) = x_1 \cdot P(x_1) + x_2 \cdot P(x_2) \\\\E(x) = 0.20 \cdot 0.60 - 0.03 \cdot 0.40 \\\\E(x) = 0.12 - 0.012 \\\\E(x) = 0.108 \\\\E(x) = 10.8\% \\\\$

Therefore, the fruit company’s expected returns are 10.8%

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3 years ago