Hannah has a chicken coop with 6 hens. Let X represent the total number of eggs the hens lay on a randomly chosen day. The distribution for X is given in the table. A 2-column table with 7 rows. Column 1 is labeled number of eggs with entries 0, 1, 2, 3, 4, 5, 6. Column 2 is labeled probability with entries 0.02, 0.03, 0.07, 0.12, 0.30, 0.28, 0.18.

What is the median of the distribution?

3
3.5
4
4.2

The correct option is;

3

Step-by-step explanation:

The data from the question is presented as follows;

Term              Number of eggs                                           Probability

1                         0                                                                         0.02

2                        1                                                                          0.03

3                        2                                                                         0.07

4 Median term  3                                                                        0.12

5                        4                                                                        0.30

6                        5                                                                        0.28

7                         6                                                                        0.18

The median of the distribution is given by the (n + 1)/2th term

Where;

n = The number of terms = 7

Therefore, the median of the distribution is the (7 + 1)/2th term which is the 4th term of the distribution, which is 3

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}