Respuesta :
(a) This is a binomial
experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
(b) Using the formula:
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
(c) Fewer than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
(b) Using the formula:
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
(c) Fewer than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
According to the flight on certain routes, the experiment is a binomial experiment and this will be determined by observing the given data.
Given :
- Flights on a certain route are on-time 80% of the time.
- 10 flights are randomly selected and the number of on-time flights is recorded.
a). There are only two possible results for each data point, a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2). Therefore it is a binomial experiment.
b). The probability that exactly 7 flights are on time
[tex]\rm P(r \;out\; of\; n) = \;^nC_r \; p^r\;q^(^n^-^r^)[/tex]
[tex]\rm P(7 \; out\; of \;10) = \; ^1^0C_7\;(0.8)^7\;(0.2)^(^1^0^-^7^)[/tex]
[tex]\rm P(7 \; out\; of \;10) = \; 0.2013[/tex]
c). Fewer than 7 flights are on time means:
[tex]\rm P(0 \; out\; of \;10) +\rm P(1 \; out\; of \;10)+\rm P(2 \; out\; of \;10)+\rm P(3 \; out\; of \;10)+\rm P(4 \; out\; of \;10)+\rm P(5 \; out\; of \;10)+\rm P(6 \; out\; of \;10) = 0.1209[/tex]
The probability that fewer than 7 flights are on time = 0.1209
d) The probability that at least 7 flights are on time = 1 - 0.1209 = 0.8791
e) The probability that between 5 and 7 flights, inclusive, are on time.
[tex]= \rm P(5 \; out\; of \;10)+\rm P(6 \; out\; of \;10) +\rm P(7 \; out\; of \;10)[/tex]
[tex]= 0.0264+0.0881+0.2013[/tex]
= 0.3158
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https://brainly.com/question/11034287
