You will need 3.333.... cups of rice to make 10 servings of casserole.
<u><em>Explanation</em></u>
The recipe which makes 6 servings of casserole, needs 2 cups of rice.
Suppose, you need cups of rice for making 10 servings of casserole.
Now, the equation according to <u>the ratio of "cups of rice" to the "servings of casserole" </u>will be.....
So, you will need 3.333.... cups of rice to make 10 servings of casserole.
Answer:
(a) Probability mass function
P(X=0) = 0.0602
P(X=1) = 0.0908
P(X=2) = 0.1704
P(X=3) = 0.2055
P(X=4) = 0.1285
P(X=5) = 0.1550
P(X=6) = 0.1427
P(X=7) = 0.0390
P(X=8) = 0.0147
NOTE: the sum of the probabilities gives 1.0068 for rounding errors. It can be divided by 1.0068 to get the adjusted values.
(b) Cumulative distribution function of X
F(X=0) = 0.0602
F(X=1) = 0.1510
F(X=2) = 0.3214
F(X=3) = 0.5269
F(X=4) = 0.6554
F(X=5) = 0.8104
F(X=6) = 0.9531
F(X=7) = 0.9921
F(X=8) = 1.0068
Step-by-step explanation:
Let X be the number of people who arrive late to the seminar, we can assess that X can take values from 0 (everybody on time) to 8 (everybody late).
<u>For X=0</u>
This happens when every couple and the singles are on time (ot).
<u>For X=1</u>
This happens when only one single arrives late. It can be #4 or #5. As the probabilities are the same (P(#4=late)=P(#5=late)), we can multiply by 2 the former probability:
<u>For X=2</u>
This happens when
1) Only one of the three couples is late, and the others cooples and singles are on time.
2) When both singles are late , and the couples are on time.
<u>For X=3</u>
This happens when
1) Only one couple (3 posibilities) and one single are late (2 posibilities). This means there are 3*2=6 combinations of this.
<u>For X=4</u>
This happens when
1) Only two couples are late. There are 3 combinations of these.
2) Only one couple and both singles are late. Only one combination of these situation.
<u>For X=5</u>
This happens when
1) Only two couples (3 combinations) and one single are late (2 combinations). There are 6 combinations.
<u>For X=6</u>
This happens when
1) Only the three couples are late (1 combination)
2) Only two couples (3 combinations) and one single (2 combinations) are late
<u>For X=7</u>
This happens when
1) Only one of the singles is on time (2 combinations)
<u>For X=8</u>
This happens when everybody is late