Week 4 Knowledge Check Homework Practice Questions
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Written Nov 25, 2025 5:42 AM - Nov 25, 2025 6:40 AMAttempt 1 of 4
- 9
Attempt Score
5%
- 9
Overall Grade (Highest Attempt)
5%
Question 1 point
Find P(Z ≤ 3). Round answer to 4 decimal places. Answer:
___0.9987___
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In Excel,
=NORM.S.DIST(3,TRUE)
Question
point
2
The size of fish is very important to commercial fishing. A study conducted in 2012
found the length of Atlantic cod caught in nets in Karlskrona to have a mean of
49.9 cm and a standard deviation of 3.74 cm (Ovegard, Berndt & Lunneryd, 2012).
Assume the length of fish is normally distributed. What is the length in cm of the
longest 15% of Atlantic cod in this area? Round answer to 2 decimal places.
Answer:
___53.78___
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Longest 15% is the upper 85%, 1 - .15 = .85
In Excel,
, =NORM.INV(0.85,49.9,3.74)
Question 3 point
Arm span is the physical measurement of the length of an individual's arms from
fingertip to fingertip. A man's arm span is approximately normally distributed with
mean of 70 inches with a standard deviation of 4.5 inches. Find length in inches of
the 99th percentile for a man's arm span. Round answer to 2 decimal places.
Answer:
___80.47___
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In Excel,
=NORM.INV(0.99,70,4.5)
Question
point
4
Which type of distribution does the graph illustrate?
Your work has been saved and submitted
Your work has been saved and submitted
Written Nov 25, 2025 5:42 AM - Nov 25, 2025 6:40 AMAttempt 1 of 4
- 9
Attempt Score
5%
- 9
Overall Grade (Highest Attempt)
5%
Question 1 point
Find P(Z ≤ 3). Round answer to 4 decimal places. Answer:
___0.9987___
Hide question 1 feedback
Feedback
In Excel,
=NORM.S.DIST(3,TRUE)
Question
point
2
The size of fish is very important to commercial fishing. A study conducted in 2012
found the length of Atlantic cod caught in nets in Karlskrona to have a mean of
49.9 cm and a standard deviation of 3.74 cm (Ovegard, Berndt & Lunneryd, 2012).
Assume the length of fish is normally distributed. What is the length in cm of the
longest 15% of Atlantic cod in this area? Round answer to 2 decimal places.
Answer:
___53.78___
Hide question 2 feedback
Feedback
Longest 15% is the upper 85%, 1 - .15 = .85
In Excel,
, =NORM.INV(0.85,49.9,3.74)
Question 3 point
Arm span is the physical measurement of the length of an individual's arms from
fingertip to fingertip. A man's arm span is approximately normally distributed with
mean of 70 inches with a standard deviation of 4.5 inches. Find length in inches of
the 99th percentile for a man's arm span. Round answer to 2 decimal places.
Answer:
___80.47___
Hide question 3 feedback
Feedback
In Excel,
=NORM.INV(0.99,70,4.5)
Question
point
4
Which type of distribution does the graph illustrate?