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creativ13 [48]
2 years ago
15

It is known that 70% of the customers in a sporting goods store purchase a pair of running shoes. a random sample of 25 customer

s is selected. assume that customers' purchases are made independently, and let x represent number of customers who purchase running shoes. find the probability distribution of x.
Mathematics
1 answer:
NikAS [45]2 years ago
8 0

Let x be the discrete random variable whose value is the number of successes in n trials.

The probability distribution function for x of the binomial distribution B(n,p) is defined as

B(n,p)= nC_xp^x(1-p)^{n-x}

Given that the random sample size is n=25

let x represent number of customers who purchase running shoes

Let "p" be the probability of customers in a sporting goods store purchase a pair of running shoes.

It is given that 70% of the customers in a sporting goods store purchase a pair of running shoes.

Thus p=\frac{70}{100}=0.7

Thus the Probability distribution of x is given by

P(X=x)= 25C_x(0.7)^x(1-0.7)^{25-x}, where x=0,1,2,3,...,25.

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It is -11.34

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You might want to stick to at most five questions at once, makes it easier for the rest of us. :)

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4 0
2 years ago
Read 2 more answers
Laurissa rolls two number cubes, each with the numbers 1 through 6. Laurissa adds the numbers that appear in the roll. Which two
defon

Answer:

Let's analyze the possible sums of both dices, i will use the notation:

Dice1 + Dice 2 = sum.

also remember that we have 2 dices, with 6 options each.

So the total number of combinations is 6*6 = 36

we have 36 possible outcomes.

I will start at the extremes, the minimum that we can sum is 2, and the maximum is 12, then:

We can have 2 if:

1 + 1 = 2

only one permutation.

and 12 if:

6 + 6 = 12

Again, only one permutation.

so 2 and 12 have the same chance (1 out of 36)

now, to have 3 we can have:

2 + 1 = 3

or

1 + 2 = 3.

and to have 11

5 + 6 = 11

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4 and 10 will have the same chance.

5 and 9 will have the same chance

6 and 8 will have the same chance

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2 years ago
Consider a disease whose presence can be identified by carrying out a blood test. Let p denote the probability that a randomly s
kifflom [539]

Answer:

The expected number of tests, E(X) = 6.00

Step-by-step explanation:

Let us denote the number of tests required by X.

In the case of 5 individuals, the possible value of x are 1, if no one has the disease, and 6, if at least one person has the disease.

To find the probability that no one has the disease, we will consider the fact that the selection is independent. Thus, only one test is necessary.

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We can then use the previously determined values to compute the expected number of tests.

E(X) = ∑x.P(X=x)

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