Both trains travel for 7 time units.
Train A travels 385 distance units, so its speed is
Train B travels 577.5 distance units, so its speed is
Note that the units are not specified, but considering that we're talking about trains, the graph is likely to be showing distance in km (or miles) and time in hours.
Also, the distance traveled by train B is not perfectly aligned with the grid, so I assumed that it was the average between the two ticks of 550 and 605.
Answer:
a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of , in which
is the mean and
is the standard deviation.
b) We have to find X when Z has a p-value of , and X is given by:
, in which
is the mean and
is the standard deviation.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
In this question:
Mean , standard deviation
a. Find the probability of a pregnancy lasting X days or longer.
The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of , in which
is the mean and
is the standard deviation.
b. If the length of pregnancy is in the lowest a%, then the baby is premature. Find the length that separates premature babies from those who are not premature.
We have to find X when Z has a p-value of , and X is given by:
, in which
is the mean and
is the standard deviation.