2. Find the probability that the flight has empty seats.
Answer:
(a) The probability of overbooking is 0.2135.
(b) The probability that the flight has empty seats is 0.4625.
Step-by-step explanation:
Let the random variable X represent the number of passengers showing up for the flight.
It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.
Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.
Number of reservations = 19
Regular customers = 14
Seats available = 17 - 14 = 3
Remaining reservations, n = 19 - 14 = 5
P (A remaining passenger will arrive), p = 0.52
The random variable X thus follows a Binomial distribution with parameters n = 5 and p = 0.52.
(1)
Compute the probability of overbooking as follows:
P (Overbooking occurs) = P(More than 3 shows up for the flight)
Thus, the probability of overbooking is 0.2135.
(2)
Compute the probability that the flight has empty seats as follows:
P (The flight has empty seats) = P (Less than 3 shows up for the flight)
Thus, the probability that the flight has empty seats is 0.4625.
Answer:
its A just took tha test
Step-by-step explanation:
The question is about representing Lola's total earnings when she works a certain number of hours. It can be answered using the equation y=18x, where y represents her total earnings and x represents the number of hours she works.
The question is about representing Lola's total earnings when she works a certain number of hours. She wrote the equation y=18x, where y represents her total earnings and x represents the number of hours she works.
This equation is similar to the example given about Svetlana's tutoring earnings, which is represented by the equation y=25+15x, where y is the total amount of money Svetlana earns and x is the number of hours she tutors.
Therefore, the subject of this question is Mathematics and it is appropriate for High School level.
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Answer:
No, the manager is not correct based on the 95% confidence interval.
Step-by-step explanation:
We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .
The Pivotal quantity for 95% confidence interval is given by;
~
where, X bar = sample mean = $1080
s = sample standard deviation = $260
n = sample size = 35 {five-week}
So, 95% confidence interval for average daily revenue, is given by;
P(-2.032 < < 2.032) = 0.95
P(-2.032 < < 2.032) = 0.95
P(-2.032 * < < 2.032 * ) = 0.95
P(X bar - 2.032 * < < X bar + 2.032 * ) = 0.95
95% confidence interval for = [ X bar - 2.032 * , X bar + 2.032 * ]
= [ 1080 - 2.032 * , 1080 + 2.032 * ]
= [ 990.70 , 1169.30 ]
No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.
Therefore, the store manager believe is not correct.
The 95% confidence interval for the store's average daily revenue is calculated to be approximately ($993.97, $1166.03). Since $1200 is outside this interval, the manager's belief that the coffee and pastry strategy will lead to an average daily revenue of $1200 is not backed by this confidence level.
In the field of statistics, a confidence interval (CI) is a type of interval estimate that is used to indicate the reliability of an estimate. The method for calculating a 95% confidence interval for the average daily revenue involves the sample mean, the standard deviation, and the z-score associated with a 95% confidence level, which is approximately 1.96. Let's use the provided data to calculate:
The range of this 95% confidence interval is from $993.97 to $1166.03. This means we are 95% confident that the true average daily revenue lies within this interval. Since $1200 lies outside this interval, the manager's belief is not supported by this confidence interval.
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Answer:
15
Step-by-step explanation:
Let f(x) be the function
A linear approximation of f is the Taylor polynomial of degree one:
Taking a = 16, and given that
we get
so
Since 16 > 14, we can deduce that this is an overestimate.