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3 years ago

Chris has a blue coat and a

Mathematics

1 answer:

Tju [1.3M]3 years ago

3 0

Answer: P = 0.125 = 1/8

Step-by-step explanation:

We know that he has a blue coat and a black coat.

If he dresses at random, then the probability of getting the blue coat is equal to the quotient between the number of blue coats (1) and the total number of coats (2).

Then the probability is:

p = 1/2

We also know that he has blue pants and brown pants, the probability of getting at random the blue pants is calculated in the same way than above, then:

q = 1/2

And for the shirt he has a blue shirt and a red one, the probability of randomly selecting the blue one is calculated in the same way than above, then:

k = 1/2

Now, the joint probability (he selects all blue clothes) is equal to the product of the individual probabilities:

P = p*q*k = (1/2)*(1/2)*(1/2) = 1/8 = 0.125

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Step-by-step explanation:

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In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi

Rina8888 [55]

Answer:

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Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n = 50

sample mean $\overline x$ = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

$X \sim N ( \mu \sigma)$

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P(X>140) = 1 - P(X ≤ 140)

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})$

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})$

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$P(X>140) = 1 - 0.6628$

$P(X>140) = 0.3372$

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

$Y \sim Binomial (np)$

$Y \sim Binomial (50,0.3372)$

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

$P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49} + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +... + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}$

P(Y ≥ 15) = 0.763

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