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katovenus [111]
3 years ago
7

When Harper commutes to work, the amount of time it takes her to arrive is normally distributed with a mean of 24 minutes and a

standard deviation of 3 minutes. Using the empirical rule, determine the interval that represents the middle 95% of her commute times.
Mathematics
1 answer:
Korvikt [17]3 years ago
7 0

Answer:

\mu -2\sigma = 24-2*3= 18

\mu -2\sigma = 24+2*3= 30

So then we can conclude that we expect the middle 95% of the values within 18 and 30 minutes for this case

Step-by-step explanation:

For this case we can define the random variable X as the amount of time it takes her to arrive to work and we know that the distribution for X is given by:

X \sim N(\mu = 24, \sigma =3)

And we want to use the empirical rule to estimate the middle 95% of her commute times. And the empirical rule states that we have 68% of the values within one deviation from the mean, 95% of the values within two deviations from the mean and 99.7 % of the values within 3 deviations from the mean. And we can find the limits on this way:

\mu -2\sigma = 24-2*3= 18

\mu -2\sigma = 24+2*3= 30

So then we can conclude that we expect the middle 95% of the values within 18 and 30 minutes for this case

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An insurance company selected samples of clients under 18 years of age and over 18 and recorded the number of accidents they had
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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

( p₁  -  p₂  )  =  0,36  -  0,25  = 0,11

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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