Respuesta :
Answer:
a. Qualitative
b. p is a random variable
c. Normal distribution
[tex]\mu_p = 0.82[/tex]
[tex]\sigma_p = 0.038[/tex]
d. [tex]P(p > 0.85) = 21.77 \%[/tex]
e. [tex]P(p < 0.75) = 3.28 \%[/tex]
Step-by-step explanation:
a. Suppose a random sample of 100 Americans is asked, "Are you satisfied with the way things are going in your life?" Is the response to this question qualitative or quantitative? Explain.
Since the question is "Are You Satisfied?" which is not a numeric value to be measured but rather satisfaction is a state of mind therefore, it is qualitative in nature.
b. Explain why the sample proportion, p, is a random variable. What is the source of the variability?
[tex]$ p = \frac{n}{N} $[/tex]
The sample proportion p represents a random variable as n number of favorable people may vary from N number of total population therefore, p is a random variable.
c. Describe the sampling distribution of p, the proportion of Americans who are satisfied with the way things are going in their life. Be sure to verify the model requirements.
The sampling distribution of p, the proportion of Americans who are satisfied with the way things are going in their life follows normal distribution if following conditions holds true.
n×p ≥ 5
n×p×(1 - p) ≥ 5
We have n = 100 and p = 0.82
100×0.82 ≥ 5
82 ≥ 5 (satisfied)
100×0.82(1 - 0.82) ≥ 5
14.76 ≥ 5 (satisfied)
The mean of the sample proportion p is
[tex]\mu_p = 0.82[/tex]
The standard deviation of the sample proportion p is
[tex]$ \sigma_p = \sqrt{\frac{p(1-p)}{n} } $[/tex]
[tex]$ \sigma_p = \sqrt{\frac{0.82(1-0.82)}{100} } $[/tex]
[tex]\sigma_p = 0.038[/tex]
d. In the sample obtained in part (a), what is the probability the proportion who are satisfied with the way things are going in their life exceeds 0.85?
[tex]P(p > 0.85) = 1 - P(p < 0.85)\\\\P(p > 0.85) = 1 - P(Z < \frac{p - \mu_p}{\sigma_p} )\\\\P(p > 0.85) = 1 - P(Z < \frac{0.85 - 0.82}{0.038} )\\\\P(p > 0.85) = 1 - P(Z < \frac{0.03}{0.038} )\\\\P(p > 0.85) = 1 - P(Z < 0.78)\\\\[/tex]
The z-score corresponding to 0.78 is 0.7823
[tex]P(p > 0.85) = 1 - 0.7823\\\\P(p > 0.85) = 0.2177\\\\P(p > 0.85) = 21.77 \%[/tex]
Therefore, there is 21.77% probability that the proportion who are satisfied with the way things are going in their life exceeds 0.85.
e. Would it be unusual for a survey of 100 Americans to reveal that 75 or fewer are satisfied with the way things are going in their life? Why?
n = 75 and N = 100
p = 75/100
p = 0.75
[tex]P(p < 0.75) = P(Z < \frac{p - \mu_p}{\sigma_p} )\\\\P(p < 0.75) = P(Z < \frac{0.75 - 0.82}{0.038} )\\\\P(p < 0.75) = P(Z < \frac{-0.07}{0.038} )\\\\P(p < 0.75) = P(Z < -1.84)\\\\[/tex]
The z-score corresponding to -1.84 is 0.03288
[tex]P(p < 0.75) = 0.03288\\\\P(p < 0.75) = 3.28 \%[/tex]
Since the probability is very low, therefore, it is indeed unusual for a survey of 100 Americans to reveal that 75 or fewer are satisfied with the way things are going in their life.
