Margin of error, e = Z*SD/Sqrt (N), where N = Sample population
Assuming a 95% confidence interval and substituting all the values;
At 95% confidence, Z = 1.96
Therefore,
0.23 = 1.96*1.9/Sqrt (N)
Sqrt (N) = 1.96*1.9/0.23
N = (1.96*1.9/0.23)^2 = 262.16 ≈ 263
Minimum sample size required is 263 students.
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).
Answer:
i)32C16
ii)1185408
<em><u>Explanation</u></em><em><u>:</u></em>
i)Total number of selected/eligible is 7+9+8+8=32
Total ways of selecting dance committee of 16 is
<em><u>3</u></em><em><u>2</u></em><em><u>C</u></em><em><u>1</u></em><em><u>6</u></em>
ii)Total ways of selecting 3 seniors from 8 is 8C3
and Total ways of selecting 6 juniors from 8 is 8C6
ways of selecting 2 sopho from 9 is 9C2
ways of selecting 5 freshman from 7 is 7C5
now, total way of selection come to be
8C3×8C6×9C2×7C5
=56×28×36×21
=1185408
✌️
Less than greater than equal Too
Answer:
Hey there!
3n-2=2n
n-2=0
n=2
Let me know if this helps :)