Question

The graph of the discrete probability to the right represents
the number of live births by a mother 40 to 44 years old
who had a live birth in 2015. Complete parts (a) through (d)
below.

0.30-
0.25-
0.20
0.15
0.10
0.05
0.00
0
0.235
1
0.270
2
0784
113 0101
-4426-0004 0.045
3
6
Number of Live Births
(a) What is the probability that a randomly selected 40- to 44-year-old mother who had a live birth in 2015 has had her fourth live birth in that year?
(Type an integer or a decimal)
(b) What is the probability that a randomly selected 40- to 44-year-old mother who had a live birth in 2015 has had her fourth or fifth live birth in that year?
(Type an integer or a decimal.)
(c) What is the probability that a randomly selected 40- to 44-year-old mother who had a live birth in 2015 has had her sixth or more live birth in that year?
(Type an integer or a decimal)
(d) If a 40-to 44-year-old mother who had a live birth in 2015 is randomly selected, how many live births would you expect the mother to have had?

Answer 1

The values of the probabilities are

• The probabilities are 0.109, 0.202, 0.106
• The expected number of births is 3

How to determine the probabilities?

The image that completes the question is added as an attachment

The probability of having her fourth live birth in that year?

From the attached graph, we have:

P(x) = 0.109 when x = 4

Hence, the probability is 0.109

The probability of having a live birth in her fourth or fifth live birth in that year?

From the attached graph, we have:

P(x) = 0.109 when x = 4

P(x) = 0.093 when x = 5

So, we have:

P(4 or 5) = 0.109 + 0.093

Evaluate

P(4 or 5) = 0.202

Hence, the probability is 0.202

The probability of having a live birth in her sixth or more live birth in that year?

This is represented as:

P(x >= 6)

From the attached graph, we have:

P(x) = 0.022 when x = 6

P(x) = 0.036 when x = 7

P(x) = 0.048 when x = 8

So, we have:

P(x >= 6) = 0.022 + 0.036 + 0.048

Evaluate

P(x >= 6) = 0.106

Hence, the probability is 0.106

How many live births would you expect the mother to have had?

This is calculated as:

$E(x) = \sum x * P(x)$

So, we have:

E(x) = 0.234 * 1 + 0.291 * 2 + 0.167 * 3 + 0.109 * 4 + 0.093 * 5 + 0.022 * 6 + 0.036 * 7 + 0.048 * 8

Evaluate

E(x) = 2.986

Approximate

E(x) = 3

Hence, the expected number of births is 3

Read more about probability at:

https://brainly.com/question/25870256

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