Hotel managers wish to learn the average length of stay of all visitors that are senior citizens. A statistician determines that
to calculate a 90% confidence level estimate of the average length of stay to within +/- 1.5 days, a total of 65 senior visitors’ check-in/check-out records will have to be examined. How many records should be looked at to obtain a 90% confidence level estimate to within +/- 0.5 days?
1 answer:
You might be interested in
Answer:
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = 1.1
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Given total number of students n(T) = 150
Given 125 of them are fluent in Swahili
Let 'S' be the event of fluent in Swahili language
n(S) = 125
The probability that the fluent in Swahili language
Let 'E' be the event of fluent in English language
n(E) = 135
The probability that the fluent in English language
n(E∩S) = 95
The probability that the fluent in English and Swahili
<u><em>Step(ii):</em></u>-
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = P(S) + P(E) - P(S∩E)
= 0.833+0.9-0.633
= 1.1
<u><em>Final answer:-</em></u>
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = 1.1