(-2,3) i believe this is correct
Answer:
The margin of error for the 99% confidence level for this sample is ±2.23.
None of the given figures is close to the answer:
E. None of the above
Step-by-step explanation:
margin of error (ME) around the mean can be calculated using the formula
ME=
where
- z is the corresponding statistic of the 99% confidence level (2.576)
- s is the population IQ standard deviation (15)
- N is the sample size (300)
Using these numbers we get:
ME=
≈ 2.23
Answer:
The correct option is (b).
Step-by-step explanation:
If X
N (µ, σ²), then
, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z
N (0, 1).
The distribution of these z-variate is known as the standard normal distribution.
The mean and standard deviation of the active minutes of students is:
<em>μ</em> = 60 minutes
<em>σ </em> = 12 minutes
Compute the <em>z</em>-score for the student being active 48 minutes as follows:

Thus, the <em>z</em>-score for the student being active 48 minutes is -1.0.
The correct option is (b).