The graph of the discrete probability to the right represents the number of live
ID: 3331433 • Letter: T
Question
The graph of the discrete probability to the right represents the number of live births by a mother 42 to 47 years old who had a live birth in 2015 0.30 0 20 0 1 Number of Live Births (a) What is the probability that a randomly selected 42-to 47-year-old mother who had a live birth in 2015 has had her fourth live birth? (Type an integer or a decimal) b) What is the probability that a randomly selected 42- to 47-year-old mother who had a live bith in 2015 has had her fourth or fitth live birth? (Type an integer or a decimal) (c) what is the probabay that a randomly selected 42-to 47-year-old mother who had a birth in 2015 has had her sixth or more live birth? (Type an integer or a decimal ) d) H a 42-to 47-year-old mother who had a ive birth In 2015 is randomly selected, how many live births would you expect the mother to have had? Round to ona decimal placa as needed ) Entor your anewor m eaoh of the enewer boxea here to searchExplanation / Answer
The probability histogram represents the number of live births by a mother 42 to 47 years old who had a live birth in 2015.
a)
The probability that a randomly selected 42- to 47-year-old mother who had a live birth in 2015 has had her fourth live birth is 0.1111.
b)
The probability that a randomly selected 42- to 47-year-old mother who had a live birth in 2015 has had her fourth or fifth live birth is 0.2101.
0.1111 + 0.099 = 0.2101
c)
The probability that a randomly selected 42- to 47-year-old mother who had a live birth in 2015 has had her sixth or more live birth is 0.103.
0.021 + 0.037 + 0.045 = 0.103
d)
This is found by computing the mean
Mean = (1 * 0.235) + (2*0.288) + (3*0.549) + (4*0.111) + (5*0.099) + (6*0.021) + (7*0.032) + (8*0.045)
= 3.009
Rounded to one decimal places is 3.0.
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