Can someone please help
2 answers:
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Answer:
There is a 90.32% probability that the cake was baked by Doug.
Step-by-step explanation:
We have these following probabilities:
A 70% probability that Doug bakes the cake.
A 30% probability that Jeremy bakes the cake.
A 40% probability that a cake baked by Doug gets a thumbs up.
A 10% that a cake baked by Jeremy gets a thumbs up.
One cake was selected at random on 10/01/2014 and got a "thumbs up".
1. Find the probability that the cake was baked by Doug.
The probability that a baked cake gets a thumbs up is:
Of those, 0.7*0.4 = 0.28 are baked by Doug.
So the probability is:
There is a 90.32% probability that the cake was baked by Doug.
Answer:
The Margin of error = 0.01
Step-by-step explanation:
<u><em>Explanation:-</em></u>
<u><em>step(i):-</em></u>
<em>Given confidence interval for the proportion of vacationers traveling abroad </em>
<em>(0.14,0.16)</em>
<em>The 95% of confidence interval for Population proportion with margin of error is determined by</em>
<em>( p⁻ - M.E , p⁻ + M.E)</em>
<u><em>step(ii):-</em></u>
<em>The margin of error is determined by</em>
<em></em><em></em>
<em>Given Confidence interval is ( 0.14 , 0.16 )</em>
Now
(( p⁻ - M.E , p⁻ + M.E) = (0.14,0.16)
Equating
p⁻ - M.E = 0.14 ...(i)
p⁻ + M.E = 0.16 ...(ii)
Solving (i) and (ii) equations , we get
p⁻ - M.E = 0.14
p⁻ + M.E = 0.16
<u> - - - </u>
- 2 M.E = -0.02
M.E = 0.01
The margin of error = 0.01
<u><em>Conclusion:-</em></u>
The margin of error = 0.01