Let x be the lengths of the steel rods and X ~ N (108.7, 0.6)
To get the probability of less than 109.1 cm, the solution is computed by:
z (109.1) = (X-mean)/standard dev
= 109.1 – 108/ 0.6
= 1.1/0.6
=1.83333, look this up in the z table.
P(x < 109.1) = P(z < 1.8333) = 0.97 or 97%
Answer:
When the range is wide
Step-by-step explanation:
Answer:
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Step-by-step explanation:
Answer: 3t-8
Step-by-step explanation:
<em>3t-8 is the answer because there are no like terms.</em>
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1
P(V|A) is not 0.95. It is opposite:
P(A|V)=0.95
From the text we can also conclude, that
P(A|∼V)=0.1
P(B|V)=0.9
P(B|∼V)=0.05
P(V)=0.01
P(∼V)=0.99
What you need to calculate and compare is P(V|A) and P(V|B)
P(V∩A)=P(A)⋅P(V|A)⇒P(V|A)=P(V∩A)P(A)
P(V∩A) means, that Joe has a virus and it is detected, so
P(V∩A)=P(V)⋅P(A|V)=0.01⋅0.95=0.0095
P(A) is sum of two options: "Joe has virus and it is detected" and "Joe has no virus, but it was mistakenly detected", therefore:
P(A)=P(V)⋅P(A|V)+P(∼V)⋅P(A|∼V)=0.01⋅0.95+0.99⋅0.1=0.1085
