According to an airline, flights on a certain route are on time 80 % of the time. suppose 10 flights are randomly selected and
the number of on-time flights is recorded. (a) explain why this is a binomial experiment. (b) find and interpret the probability that exactly 7 flights are on time. (c) find and interpret the probability that fewer than 7 flights are on time. (d) find and interpret the probability that at least 7 flights are on time. (e) find and interpret the probability that between 5 and 7 flights, inclusive, are on time.
<span>(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:</span><span> 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. </span>
