The answer to your question is x=8
Answer:
The 95% confidence interval for the proportion of students who get coaching on the SAT is (0.1232, 0.147).
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 z-score that has a p-value of .
427 had paid for coaching courses and the remaining 2733 had not.
This means that
95% confidence level
So , z is the value of Z that has a p-value of , so .
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the proportion of students who get coaching on the SAT is (0.1232, 0.147).
Answer:
Step-by-step explanation:
X no of questions student answers is binomial with n =40 and p =0.5
If approximated to normal, X is Normal with
mean = np = 20 and variance = npq = 10
P(X>N) >0.10
We use std normal distribution table to get z value first then convert to x value
So
This is with continuity correction.
Hence without continuity correction this equals 7.2-0.5 = 6.7
x>6.7
n = 7
A translation of right 3 up 1