Answer:
a. Average shear strength = 385.3 ; standard deviation of shear strength is ±19.25
b. At 95%, X = 1.645σ + μ
c. P(X ≤ 400) = 27.77% = 0.2777
Step-by-step explanation:
Mean = (409 + 378 + 358 + 362 + 389 + 404 + 415 + 375 + 367 + 396)/10
Mean = 3853/10
Mean = 385.3
Therefore, Average shear strength = 385.3
Variance = (409 - 385.3)² + (378 - 385.3)²+ (358 - 385.3)² + (362 - 385.3)²+ (389 - 385.3)²+ (404 - 385.3)²+ (415 - 385.3)²+ (375 - 385.3)²+ (367 - 385.3)²+ (396 - 385.3)²/10
Variance = 3704.1/10 = 370.41
Standard deviation = √370.41 = 19.25
Therefore, standard deviation of shear strength is ±19.25
b. Using normal distribution table at 95%, z = 1.645
z= (X - μ)/σ
Therefore X = 1.645σ + μ
c. X = 400; μ= 385.3 ; σ = 19.25
z= (X - μ)/σ
z= (400 - 385.3)/19.25
z = 0.764
Using the normal distribution table, P = 0.2777
P(X ≤ 400) = 27.77%
