Suppose 65 percent of the populatio n of a large community i

Suppose 65 percent of the populatio n of a large community is in favor of a new proposition. A random sample of 100 people is selected from the community. a) What is the (exact) distribution of X, the number of people among the 100 who are in favor of the proposition? Remember to provide the parameter(s). b) What is the approximate distribution of X? Remember to provide the parameter(s) and check the appropriate condition(s) for the approximation. c) Use the distribution in b) to approximate the probability that among the 100 people, i) at least 50 are in favor of the proposition, ii) between 60 and 70 inclusive who are in favor, iii) less than 75 are in favor?

Solution

a)

It is a binomial distribution, with n = 100, p = 0.65.

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b)

Here,

np = 65 > 10
n(1-p) = 35 > 10

As both quantities are greater than 10, then we can to the normal approximation.

Here, the mean and standard deviations are

u = mean = np =    65 [ANSWER]      
          
s = standard deviation = sqrt(np(1-p)) =    4.769696007   [ANSWER]  

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c)

i)

We first get the z score for the critical value:          
          
x = critical value =    49.5      
u = mean = np =    65      
          
s = standard deviation = sqrt(np(1-p)) =    4.769696007      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -3.249682994      
          
Thus, the left tailed area is          
          
P(z >   -3.249682994   ) =    0.999422331 [ANSWER]

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ii)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    59.5      
x2 = upper bound =    70.5      
u = mean = np =    65      
          
s = standard deviation = sqrt(np(1-p)) =    4.769696007      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.15311332      
z2 = upper z score = (x2 - u) / s =    1.15311332      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.124431937      
P(z < z2) =    0.875568063      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.751136125   [ANSWER]

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iii)

We first get the z score for the critical value:          
          
x = critical value =    74.5      
u = mean = np =    65      
          
s = standard deviation = sqrt(np(1-p)) =    4.769696007      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    1.99174119      
          
Thus, the left tailed area is          
          
P(z <   1.99174119   ) =    0.976800269 [ANSWER]