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1. See Video Solution
2.
a. About 85% of all the past calls were for medical assistance.
b. P(call is not for medical assistance) = 1 – 0.85 = 0.15.
c. P(two successive calls are both for medical assistance) = (0.85)(0.85) = 0.7225.
d. P(first call is for medical assistance and second call is not for medical assistance)
= (0.85)(0.15) = 0.1275
e. P(exactly one of two calls is for medical assistance) = P(first call is for medical
assistance and the second is not) + P(first call is not for medical assistance but the
second is) = (0.85)(0.15) + (0.15)(0.85) = 0.255.
f. Probably not. There are likely to be several calls related to the same event -
several reports of the same accident or fire that would be received close together
in time.
3. (“ ” “ ” “ ”) . / . / . /
4. See Video Solution
5.
a. The "expert" assumed that the positions of the two valves were independent.
b. The position of the two valves is not independent but rather dependent. The
effect of the error makes the probability much smaller. The actual probability is
compared to .
6.
a. Assuming that whether Jeanie forgets to do one of her “to do” list items is
independent of whether or not she forgets any other of her “to do” list items, the
probability that she forgets all three errands = (0.1)(0.1)(0.1) = 0.001.
b. ( )
( )
c. P(remembers the first errand, but not the second or the third) = (0.9)(0.1)(0.1) =
0.009.
5.1 The Ideas Behind the Continuous Distribution
1.
a.
Statistics for Decision-Making in Business © Milos Podmanik Page 230