The randomly selected subject is given bone density test which are normally distributed with mean =0 and standard deviation =1. It means bone density test score is standard normal variable.
The probability that a given score is less than 2.23
P(z < 2.23)
Using z score table to find probability below z=2.23
P(z < 2.23) = 0.9871
The probability that a given score is less than 2.23 is 0.9871
Answer:
The 95% confidence interval for the proportion of young adults with pierced tongues who have receding gums is (0.24, 0.506).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
For this problem, we have that:
95% confidence level
So , z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the proportion of young adults with pierced tongues who have receding gums is (0.24, 0.506).