Answer:
c = ≥ (23 - 15) therefore c = ≥ 8
Area of STAR is 12, Area of SKCO= 10.5
Step-by-step explanation:
(a) 0.7967
(b) 0.6826
(c) 0.3707
(d) 0.9525
(e) 0.1587
The random variable <em>X</em> follows a Normal distribution with mean <em>μ</em> = 10 and variance <em>σ</em>² = 36.
(a)
Compute the value of P (X > 5) as follows:
Thus, the value of P (X > 5) is 0.7967.
(b)
Compute the value of P (4 < X < 16) as follows:
Thus, the value of P (4 < X < 16) is 0.6826.
(c)
Compute the value of P (X < 8) as follows:
Thus, the value of P (X < 8) is 0.3707.
(d)
Compute the value of P (X < 20) as follows:
Thus, the value of P (X < 20) is 0.9525.
(e)
Compute the value of P (X > 16) as follows:
Thus, the value of P (X > 16) is 0.1587.
**Use a <em>z</em>-table for the probabilities.
#1-List all outcomes for choosing the digit. *
0)
O
A: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
B: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
C: 1, 2, 3, 4, 5, 6, 7, 8, 9
OD: a number
Both B and C
Answer: 17.75
The interquartile range(IQR) is the 3rd quartile - the 1st quartile.
How to get quartiles:
First get the median:
3.5, 10.4, 16, 21.7, 27.7
10.4, 16, 21.7
16
Then find the median of the first half of data(3.5, 10.4)
(3.5+10.4)/2 = 6.95
Then find the median of the last half of data(21.7, 27.7)
(21.7+27.7)/2 = 24.7
Then to get the IQR subtract 6.95 from 24.7 to get 17.75
Hope it helps <3