Answer:
a) z = 1.16
b) P(x ≥ 2) = 0.8508
c) z = 1.555
Step-by-step explanation:
a) Find z such that 87.7% of the standard normal curve lies to the left of z.
Lies to the left, so it has a pvalue of 0.877. This is z = 1.16
b) Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
μ = 2.4; σ = 0.35
P(x ≥ 2) =
This is 1 subtracted by the pvalue of Z when X = 2, in which
has a pvalue of 0.1492.
So
P(x ≥ 2) = 1 - 0.1492 = 0.8508.
c) Find z such that 6% of the standard normal curve lies to the right of z.
To the right, so z has a pvalue of 1-0.06 = 0.94%. So z = 1.555.
Answer:
m = 1/3x
Step-by-step explanation:
slope = y^2 - y^1 / x^2 - x^1
m = 4 - 5 / 3 - 6
m = -1 / -3
m = 1/3
m = 1/3x --> the slope
Part A)
Focus on the first column where it says "test finds student lying" since it asks "what percentage..were found to be lying". We can ignore the second column completely.
We have 282 people found to be lying based on the test (bottom of the first column). Of this total, 200 have not cheated meaning that the student isn't lying. This is based on the overall premise given at the top that states "All of them denied having cheated"
Simply divide the two values: 200/282 = 0.70921985815602 which rounds to roughly 0.7092 if you round to four decimal places
Now move the decimal over 2 spots to the right to convert to a percentage
0.7092 ---> 70.92%
Answer: 70.92%
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Part B)
We'll use the result from part A. Subtract it from 100%
100% - 70.92% = (100 - 70.92)% = 29.08%
Answer: 29.08%
Note: another way to get this answer is to divide the values 82 over 282 to get 82/282 = 0.29078014184398 and we get the same answer roughly (after rounding)
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Part C)
Now we focus on the second column only. We have 1718 people total and of this total, 12 have cheated. Since they initially deny cheating but they really did cheat, this is a lie they tell.
So 12/1718 = 0.0069848661234 which rounds to 0.0070 and that converts to 0.7%
Note: this isn't 7% or 70%, its point 7 percent
Answer: 0.7%
