=687.33.
Let D = “a home has a desktop computer”; L = “a home has a laptop computer.” We are given that P
(D) = 0.8 and P (D and L) = 0.3. Thus,
- The situation can be pictured as shown below. The shaded area is a trapezoid whose area is
- The fact that Harv’s pumpkin is at the 90th percentile means that it is larger than 90% of the
pumpkins in the contest. From the table of Standard Normal Probabilities, the area to the left of a term
with a z -score of 1.28 is about 0.90. Thus,
So, Harv’s pumpkin weighed about 148 pounds (for your information, the winning pumpkin at the
Half Moon Bay Pumpkin Festival in 2008 weighed over 1528 pounds!). Seven of the 62 pumpkins
weighed more than 1000 pounds!
Since , we have P (X = 4) = 1 – P (X = 2) – P (X = 3) = 1 – 0.3 – 0.5 = 0.2. Thus, filling in the
table for X , we have
Since X and Y are independent, P (X = 4 and Y = 3) = P (X = 4) · P (X = 3). We are given that P (X =
4 and Y = 3) = 0.03. Thus, P (X = 4) · P (Y = 3) = 0.03. Since we now know that P (X = 4) = 0.2, we
have (0.2). P (Y = 3) = 0.03, which gives us .Now, since ΣP (Y ) = 1, we have P (Y = 5) = 1 – P (Y = 3) – P (Y = 4) – P (Y = 6) = 1 – 0.15 – 0.1 –
0.4 = 0.35.
Let A = “the number is divisible by 24” = {24, 48, 72, 96}. Let B = “the number is divisible by 36”
= {36, 72}.
Note that P (A and B) = (72 is the only number divisible by both 24 and 36 ).