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3 years ago

C=100C=100\pi (P)=3.14π(P)=3.14D=C\div \piD=C÷πD=?D=?

Mathematics

1 answer:

Vsevolod [243]3 years ago

5 0

31.85 if it’s rounded

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Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, of all homeowners are insured

ANTONII [103]

Missing part of the question

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 25% of all homeowners.....................

Answer:

$P(X = x) = ^4C_x * 0.25^x * 0.75^{4-x}$

$Mean = 1$

$Pr = 0.2617$

Step-by-step explanation:

Given

$n = 4$

$p = 25\%$

Solving (a): The probability distribution of x

We have:

$n = 4$

$p = 25\%$ $= 0.25$

The probability of not having earthquake insurance (q) is:

$q = 1 - p$

$q = 1 - 0.25$ $= 0.75$

If x has insurance, then n - x do not.

The distribution follows a binomial pattern. So, the probability distribution is:

$P(X = x) = ^nC_x * 0.25^x * 0.75^{n-x}$

Substitute 4 for n

$P(X = x) = ^4C_x * 0.25^x * 0.75^{4-x}$

Solving (b): The most likely value of x i.e. The mean

We have:

$n = 4$ and

$p = 25\%$ $= 0.25$

$Mean = np$

$Mean = 4 * 0.25$

$Mean = 1$

Solving (c): At least 2 of 4 selected have earthquake insurance.

This is calculated as

$Pr = P(x = 2) + P(x = 3) + P(x = 4)$

$P(X = x) = ^4C_x * 0.25^x * 0.75^{4-x}$

$P(X=2) = ^4C_2 * 0.25^2 * 0.75^{4-2}$

$P(X=2) = 6 * 0.25^2 * 0.75^2 = 0.2109375$

$P(X=3) = ^4C_3 * 0.25^3 * 0.75^{4-3}$

$P(X=3) = 4 * 0.25^3 * 0.75 = 0.046875$

$P(X=4) = ^4C_4 * 0.25^4 * 0.75^{4-4}$

$P(X=4) = 1 * 0.25^4 * 1= 0.00390625$

So:

$Pr = P(x = 2) + P(x = 3) + P(x = 4)$

$Pr = 0.2109375 + 0.046875 + 0.00390625$

$Pr = 0.26171875$

$Pr = 0.2617$ --- approximated

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