A) Variance = 10.24
B) Standard Deviation = 3.2
One of the measurements of dispersion is the standard deviation, which is exclusively used for quantitative data. It aids in determining if the data's mean is a suitable measurement to reflect the core value.
TIME FREQUENCY(f) MIDPOINT(x) d d² fd fd²
0 - 0.9 43 0.45 -3 9 -129 387
1.0 - 1.9 17 1.45 -2 4 -34 68
2.0 - 2.9 19 2.45 -1 1 -19 19
3.0 - 3.9 18 3.45 0 0 0 0
4.0 - 4.9 14 4.45 1 1 14 14
5.0 -5.9 16 5.45 2 4 32 64
∑f = 127
∑fd = -136
∑fd² = 552

Standard Deviation = 

√4.35 - √1.15
Standard Deviation = 3.2
(SD)² = (3.2)² = 10.24
Variance = 10.24
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Answer:
Equation of the circle (x-3)²+(y-5)²=(6.4)²
x² -6x +9 +y² -10y +25 = 40.96
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given endpoints of diameter P(−2, 1) and Q(8, 9)
Centre of circle = midpoint of diameter
Centre = 
Centre (h, k) = (3 , 5)
<u><em>Step(ii):-</em></u>
The distance of two end points
PQ = 

PQ = √164 = 12.8
Diameter d = 2r
radius r = d/2
Radius r = 6.4
<u><em>Final answer:-</em></u>
Equation of the circle
(x-h)²+(y-k)² = r²
(x-3)²+(y-5)²=(6.4)²
x² -6x +9 +y² -10y +25 = 40.96
x² -6x +y² -10y = 40.96-34
x² -6x +y² -10y -7= 0