0.217
Yes quite unusual ( See explanation )
Step-by-step explanation:
Solution:-
- Denote among population ( P ) of the proportion of Americans who are satisfied with flow of their lives are p = 0.82.
- A random sample of n = 100 Americans were asked whether they are satisfied with the way things are going in your life.
- The sampling distribution of population proportion will be checked for normality as follows:
n = 100 , p^ = 0.82
- The requirements of normal distribution of sample proportion distribution:
n*p^ & n( 1 - p^ ) ≥ 10
n*p^*( 1 - p^ ) ≥ 10
- Verify the assumption of sample proportion to be normally distributed:
n*p^ = 100*0.82 = 82 ≥ 10
n*( 1 - p^ ) = 100*( 1 - 0.82 ) = 18 ≥ 10
n*p*( 1 - p^ ) = 100*0.82*( 1 - 0.82 ) = 14.76 ≥ 10
- Therefore, we can assume the sample proportion to be normally distributed as per Central Limit Theorem i.e sample is sufficiently large.
- Normal distribution of population proportion ( P ) is defined by two parameters that are defined as:
P ~ Norm ( μ_p , σ_p^2 )
Where,
μ_p: The mean of proportion
σ_p: The standard deviation of proportion
- The parameters for normally distributed sample proportion will be estimated using point estimation and point standard error estimation of sample population distribution.
Estimation:
μ_p = sample proportion = p^ = 0.82
σ_p = 
= 0.03842
Hence,
P ~ Norm ( 0.82 , 0.03842^2 )
- The probability of proportion of people who answered yes to the survey question is greater than 0.85.
- We are to compute the probability of p ( P > 0.85 ) using the normal sample distribution of ( P ) determined above:
- The Z-score for the limiting value is:
Z = ( P - μ_p ) / σ_p
Z = ( 0.85 - 0.82 ) / 0.03842
Z = 0.78084
- Now using standard normal tables or normal calculator determine the corresponding probability:
p ( P > 0.85 ) = p ( Z > 0.78084 )
= 0.217439
The probability that at-least 85 people in the random sample of n = 100 Americans who answered " yes " to the survey question is 0.217.
- If x = 75 or fewer Americans say that they are satisfied with the way things are going in their life among a population of N = 100.
- The proportion ( P ) of the Americans that answered " yes " to the survey question:
P = x / N
P = 75 / 100 = 0.75
- Now we will compare the population proportion ( P ≤ 0.75 ) against the sampling distribution of ( P ). Compute the Z-score of the limiting value:
Z = ( P - μ_p ) / σ_p
Z = ( 0.75 - 0.82 ) / 0.03842
Z = -1.82196
- Now using standard normal tables or normal calculator determine the corresponding probability:
p ( P ≤ 0.75 ) = p ( Z ≤ -1.82196 )
= 0.03422543
This means that about 3.4 out of 100 random samples of size 100 will result in 75% or lower are satisfied with their life if the population proportion of Americans who are satisfied with their life is 0.82. Hence, p ( P < 0.75 ) = 3 is quite unusual from a survey of 100 Americans to show that less than 75% answered yes to the posed question.
