Question 5 (1 point) Fifty soil samples from a construction site were tested for shear strength The mean of the samples was determined to be 3785 kN / m2 The standard deviation was estimated to be 440 kN / m2 Assuming the data are Normally distributed what is the probability of observing a shear strength less than 2961 Your Answer

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StudyX AI Fast Model · Accepted Answer

Solution By Steps
Step 1: Z-Score Calculation 
Calculate the Z-score using the formula: $ Z = \frac{X - \mu}{\sigma} $, where $ X $ is the value (2,961 kN/m²), $ \mu $ is the mean (3,785 kN/m²), and $ \sigma $ is the standard deviation (440 kN/m²).

Step 2: Z-Score Interpretation 
Interpret the Z-score obtained in the previous step to find the corresponding probability from the standard normal distribution table.

Step 3: Probability Calculation 
Find the probability of observing a shear strength less than 2,961 kN/m² using the standard normal distribution table.

Final Answer
The probability of observing a shear strength less than 2,961 kN/m² is approximately 0.1587.

Key Concept
Normal Distribution

Key Concept Explanation
In a normal distribution, the probability of a random variable falling below a certain value can be calculated using Z-scores and the standard normal distribution table. This concept is crucial in various fields, including statistics, engineering, and natural sciences, to analyze and interpret data distributions.

Question 5 1 point Fifty soil samples from a construction site were tested for shear strength The mean of the samples was determined to be 3785 kN m2 The standard deviation was estimated to be 440 kN...

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Question 5 1 point Fifty soil samples from a construction site were tested for shear strength The mean of the samples was determined to be 3785 kN m2 The standard deviation was estimated to be 440 kN...

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Answer & Explanation

Follow-up Knowledge or Question

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