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BartSMP [9]
3 years ago
10

The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen

dent.
a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).
b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).
c. If you call 22 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.
Mathematics
1 answer:
vfiekz [6]3 years ago
7 0

Answer:

a) 0.2581

b) 0.4148

c) 17

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

p = 0.75

a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is P(X = 9) when n = 12. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 9) = C_{12,9}.(0.75)^{9}.(0.25)^{3} = 0.2581

b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is P(X \geq 16) when n = 20

So

P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 16) = C_{20,16}.(0.75)^{16}.(0.25)^{4} = 0.1897

P(X = 17) = C_{20,17}.(0.75)^{17}.(0.25)^{3} = 0.1339

P(X = 18) = C_{20,18}.(0.75)^{18}.(0.25)^{2} = 0.0669

P(X = 19) = C_{20,19}.(0.75)^{19}.(0.25)^{1} = 0.0211

P(X = 20) = C_{20,20}.(0.75)^{20}.(0.25)^{0} = 0.0032

So

P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1897 + 0.1339 + 0.0669 + 0.0211 + 0.0032 = 0.4148

c. If you call 22 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.

The expected value of the binomial distribution is:

E(X) = np

In this question, we have n = 22

So

E(X) = 22*0.75 = 16.5

The closest integer to 16.5 is 17.

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Step-by-step explanation:

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3 years ago
An insurance company selected samples of clients under 18 years of age and over 18 and recorded the number of accidents they had
Stolb23 [73]

Answer:

Q1 z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the that means have no difference

Q2  CI 95 %  =  (  0,056 ;  0,164 )

Step-by-step explanation:

Sample information for people under 18

n₁  =  500

x₁ =  180

p₁  =  180/ 500    p₁  =  0,36    then  q₁  =  1 -  p₁     q₁ =  0,64

Sample information for people over 18

n₂  =  600

x₂  =  150

p₂  =  150 / 600   p₂ =  0,25   then   q₂  =  1 - p₂   q₂ =  1 - 0,25   q₂ = 0,75

Hypothesis Test

Null hypothesis                        H₀              p₁  =  p₂

Alternative Hypothesis           Hₐ              p₁  ≠  p₂

The alternative hypothesis indicates that the test is a two-tail test.

We will use the approximation to normal distribution of the binomial distribution according to the sizes of both samples.

Testin at CI =  95 %    significance level is  α = 5 %   α  =  0,05  and

α/ 2  =  0,025   z (c) for that α  is from z-table:

z(c) = 1,96

To calculate   z(s)

z(s)  =  ( p₁  -   p₂ ) / EED

EED = √(p₁*q₁)n₁  +  (p₂*q₂)/n₂

EED = √( 0,36*0.64)/500  +  (0,25*0,75)/600

EED = √0,00046  +  0,0003125

EED = 0,028

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Then

z(s)  =  0,11 / 0,028

z(s) = 3,93

Comparing  z(s) and  z (c)    z(s) > z(c)

z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the idea of equals means

Q2  CI  95 %   =  (  p₁  -  p₂  ) ±  z(c) * EED

CI 95%  =  ( 0,11   ±  1,96 * 0,028 )

CI 95%  = (  0,11  ±  0,054 )

CI 95 %  =  (  0,056 ;  0,164 )

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