Paige can run one lap around a track in 3 1/5 minutes. How long would it take her to run 6 1/2 laps?
1 answer:
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Answer:
a: 28 < µ < 34
Step-by-step explanation:
We need the mean, var, and standard deviation for the data set. See first attached photo for calculations for these...
We get a mean of 222/7 = 31.7143
and a sample standard deviation of: 4.3079
We can now construct our confidence interval. See the second attached photo for the construction steps.
They want a 90% confidence interval. Our sample size is 7, so since n < 30, we will use a t-score. Look up the value under the 10% area in 2 tails column, and degree of freedom is 6 (degree of freedom is always 1 less than sample size for confidence intervals when n < 30)
The t-value is: 1.943
We rounded down to the nearest person in the interval because we don't want to over estimate. It said 28.55, so more than 28 but not quite 29, so if we use 29 as the lower limit, we could over estimate. It's better to use 28 and underestimate a little when considering customer flow.
Answer:
0.335
Step-by-step explanation:
1. There is a 30 percent chance of a flight being delayed because of icy weather ,then the probability of being delayed is 0.3 and of being not delayed is 0.7.
2. If a flight is delayed because of icy weather, there is a 10 percent chance the flight will also be delayed because of a mechanical problem, then the probability of being delayed is 0.1 and the probabilty of not being delayed is 0.9.
3. If a flight is not delayed because of icy weather, there is a 5 percent chance that it will be delayed because of a mechanical problem (MP), then the probability of being delayed is 0.05 and the probabilty of not being delayed is 0.95. (See attached probability tree)
Delayed of icy weather - 0.3
Delayed of MP when weather is not icy - 0.7·0.05=0.035
Now, if one flight is selected at random from the airport in January, the probability that the flight selected will have at least one of the two types of delays is
0.3+0.035=0.335
