Question

The graph of the discrete probability to the right represents the number of live births by a mother 43 to 47 years old who had a live birth in 2015. Complete parts (a) through (d) below.

0123456780.000.050.100.150.200.250.30Number of Live BirthsProbability 0.239 0.289 0.165 0.114 0.093 0.018 0.038 0.044 A histogram with a horizontal axis labeled “Child” from 1 to 8 in intervals of 1 and a vertical axis labeled "Probability" from 0 to 0.30 in intervals of 0.05 has eight vertical bars of width 1, centered along the horizontal axis labels, with labeled heights. The heights of the vertical bars are as follows, with the horizontal axis label listed first and the bar height listed second: 1, 0.239; 2, 0.289; 3, 0.165; 4, 0.114; 5, 0.093; 6, 0.018; 7, 0.038; 8, 0.044.

(a) What is the probability that a randomly selected

43-

to

47-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

43-

to

47-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

43-

to

47-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

43-

to

47-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?

..............??????? (Round to one decimal place as needed.)

Answer 1

(a)

Probability that a randomly selected 43 -to 47 -year-old mother who had a live birth in 2015 has had her fourth live birth in that year = P(X = 4)

= 0.109

(b)

Probability that a randomly selected 43 -to 47 -year-old mother who had a live birth in 2015 has had her fourth or fifth live birth in that year = P(X = 4 or 5)

P(X = 4) + P(X = 5) = 0.109 + 0.096 = 0.205

(c)

Probability that a randomly selected 43 -to 47 -year-old mother who had a live birth in 2015 has had six or more live birth in that year = P(X >= 6)

P(X = 6) + P(X = 7) + P(X = 8) = 0.017 + 0.032 + 0.048 = 0.097

(d)

Expected number of live births = E(X)

= 0.234 * 1 + 0.299 * 2 + 0.165 * 3 + 0.109 * 4 + 0.095 * 5 + 0.017 * 6 + 0.032 * 7 + 0.048 * 8

= 2.953 3.0