What's the correct equation?
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Let Z be the reading on thermometer. Z follows Standard Normal distribution with mean μ =0 and standard deviation σ=1
The probability that randomly selected thermometer reads greater than 2.07 is
P(z > 2.07) = 1 -P(z < 2.07)
Using z score table to find probability below z=2.07
P(Z < 2.07) = 0.9808
P(z > 2.07) = 1- 0.9808
P(z > 2.07) = 0.0192
The probability that a randomly selected thermometer reads greater than 2.07 is 0.0192
Perhaps you mean to find the volume under 
? It seems like you have to rely on calculus otherwise (in the case that you indeed mean 
).
Assuming this, note that describes the top half of a sphere with radius 8, which means the volume below this surface over the disk 
is simply half the volume of the sphere. Thus the volume is
Assuming this, note that
The rest of the question is the attached figure.
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Δ AYW a right triangle at Y ⇒⇒⇒ ∴ WA² = AY² + YW²
And AY = YB ⇒⇒⇒ ∴ WA² = YB² + YW² → (1)
Δ BYW a right triangle at Y ⇒⇒⇒ ∴ WB² = BY² + YW² → (2)
From (1) , (2) ⇒⇒⇒ ∴ WA = WB →→ (3)
Δ CXW a right triangle at Y ⇒⇒⇒ ∴ WC² = CX² + XW²
And CX = XB ⇒⇒⇒ ∴ WC² = XB² + XW² → (4)
Δ BXW a right triangle at Y ⇒⇒⇒ ∴ WB² = XB² + XW² → (5)
From (4) , (5) ⇒⇒⇒ ∴ WC = WB →→ (6)
From (3) , (6)
WA = WB = WC
given ⇒⇒⇒ WA = 5x – 8 and WC = 3x + 2
∴ <span> 5x – 8 = 3x + 2</span>
Solve for x ⇒⇒⇒ ∴ x = 5
∴ WB = WA = WC = 3*5 + 2 = 17
The correct answer is option D. WB = 17
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Δ AYW a right triangle at Y ⇒⇒⇒ ∴ WA² = AY² + YW²
And AY = YB ⇒⇒⇒ ∴ WA² = YB² + YW² → (1)
Δ BYW a right triangle at Y ⇒⇒⇒ ∴ WB² = BY² + YW² → (2)
From (1) , (2) ⇒⇒⇒ ∴ WA = WB →→ (3)
Δ CXW a right triangle at Y ⇒⇒⇒ ∴ WC² = CX² + XW²
And CX = XB ⇒⇒⇒ ∴ WC² = XB² + XW² → (4)
Δ BXW a right triangle at Y ⇒⇒⇒ ∴ WB² = XB² + XW² → (5)
From (4) , (5) ⇒⇒⇒ ∴ WC = WB →→ (6)
From (3) , (6)
WA = WB = WC
given ⇒⇒⇒ WA = 5x – 8 and WC = 3x + 2
∴ <span> 5x – 8 = 3x + 2</span>
Solve for x ⇒⇒⇒ ∴ x = 5
∴ WB = WA = WC = 3*5 + 2 = 17
The correct answer is option D. WB = 17