The graph of the discrete probability to the right represents
the number of live births by a mother 43 to47 years old who had a live birth in2015. 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.)
(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
The graph of the discrete probability to the right representsthe number of live births by...
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32.) the data on the right represent the number of live multiple delivery births (3 or more babies) in a particular year for women 15 to 54 years old. use data to complete parts a through d The data on the right represent the number of the multiple delivery birthe (three or more babies) na particular you for women 16 to 5 years old. Use the date to complete parts (4) through (4) below Number of Multiple Birth Age 15-19...
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