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natali 33 [55]
3 years ago
13

The probability that a randomly selected individual in a certain community has made an online purchase is 0.41 . Suppose that a

sample of 13 people from the community is selected, what is the probability that at most 3 of them has made an online purchase?
Write only a number as your answer. Round to 2 decimal places (for example 0.24). Do not write as a percentage.
Mathematics
1 answer:
andreev551 [17]3 years ago
4 0

Answer:

0.15

Step-by-step explanation:

What we can use to solve this problem is the Bernoulli approximation of the Binomial distribution.

When we say at most 3, it means 0 or 1 or 2 or 3 people out of the 13 made the purchase

Mathematically, the Bernoulli approximation is applied as follows

P(X = n) = nCr p^r q^n-r

let p = probability of making online purchase = 0.41

while q = 1-p = 1-0.41 = 0.59 which is the probability of not making an online purchase.

Mathematically for each of the approximations up to 3, we have ;

13C0 P^0 q^13 + 13C1 P^1 q^12 + 13C2 P^2 q^11 + 13C3 P^3 q^10

Making substitutions, we have

{1 * 0.41^0 * 0.59^13} + {13 * 0.41^1 * 0.59^12} + {78 * 0.41^2 * 0.59^11} + {286 * 0.41^3 * 0.59^10}

= 0.150820944904 which is 0.15 to 2 decimal places

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Answer:

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Step-by-step explanation:

Normal price

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3 years ago
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scZoUnD [109]

The answer is 2 1/2. Because you need to cross cancel then you take the "freeway", and you get 5/2 and simplified is 2 1/2.And since half equals .5 it would be 2.5 just in case you also needed decimal.

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3 years ago
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KIM [24]

Answer:

B.  $136,960.54

Step-by-step explanation:

Rate = r = 4%

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No. pf Years = n = 21

Present Value = P = ?

A = P [1 + (r / b)]ⁿᵇ

$315,930 = P [1 + (0.04 / 4]²¹ ˣ ⁴

$315,930 = P [1 + 0.01]⁸⁴

$315,930 = P [1.01]⁸⁴

$315,930 = P x 2.306723

P = $315,930 / 2.306723

P = $136,960.54

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