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otez555 [7]
2 years ago
15

A sample of 1500 people from a certain industrial community showed that 800 people suffered from a lung disease. A sample of 100

0 people from a rural community showed that 300 suffered from the same lung disease. Assume these two proportions give good estimates of population proportions, respectively. Find the probability that the difference of the proportions (industrial - rural) is greater than 0.2 if two samples of size 150 and 100 are drawn from the two populations, respectively.
Mathematics
1 answer:
ss7ja [257]2 years ago
3 0

Answer:

The probability that the difference of the proportions (industrial - rural) is greater than 0.2 is 0.7054

Step-by-step explanation:

Solution

Given that:

n₁ = 1500

x₁ = 800

p₁ =800/1500 =0.533

q₁ = 1- 0.533 = 0.467

Thus,

n₂ =1000

x₂ = 300

p₂ = 300/1000 = 0.3

q₂ = 1-0.3 =0.7

So,

p₁ - p₂ = 0.533 - 0.3

=0.233

Now,

SE (p₁ - p₂ ) =√0.533 * 0467/150 + 0.3 * 0.7/1000

=0.0613

So,

p ( p₁ - p₂  > 0.2

= p ( Ƶ > 0.2 - 0.233/0.0613

p = ( Ƶ > - 0.54)

= 0.7054

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3 years ago
Do you tailgate the car in front of you? About 35% of all drivers will tailgate before passing, thinking they can make the car i
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Answer:

(a) The histogram is shown below.

(b) E (X) = 4.2

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Step-by-step explanation:

Let <em>X</em> = <em>r</em><em> </em>= a driver will tailgate the car in front of him before passing.

The probability that a driver will tailgate the car in front of him before passing is, P (X) = <em>p</em> = 0.35.

The sample selected is of size <em>n</em> = 12.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 12 and <em>p</em> = 0.35.

The probability function of a binomial random variable is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x}

(a)

For <em>X</em> = 0 the probability is:

P(X=0)={12\choose 0}(0.35)^{0}(1-0.35)^{12-0}=0.006

For <em>X</em> = 1 the probability is:

P(X=1)={12\choose 1}(0.35)^{1}(1-0.35)^{12-1}=0.037

For <em>X</em> = 2 the probability is:

P(X=2)={12\choose 2}(0.35)^{2}(1-0.35)^{12-2}=0.109

Similarly the remaining probabilities will be computed.

The probability distribution table is shown below.

The histogram is also shown below.

(b)

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E(X)=np

The expected number of vehicles out of 12 that will tailgate is:

E(X)=np=12\times0.35=4.2

Thus, the expected number of vehicles out of 12 that will tailgate is 4.2.

(c)

The standard deviation of a Binomial distribution is:

SD(X)=np(1-p)

The standard deviation of vehicles out of 12 that will tailgate is:

SD(X)=np(1-p)=12\times0.35\times(1-0.35)=2.73\\

Thus, the standard deviation of vehicles out of 12 that will tailgate is 2.73.

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3 years ago
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