Question

The data on the table represent the number of live multiple delivery births (three or more babies) in a particular year...

The data on the table represent the number of live multiple delivery births (three or more babies) in a particular year for women 15 to 54 years old. Use the data to determine the following. (Round all answers three decimals)

Determine the probability that a randomly selected multiple birth for women 15-54 years old involved a mother 30 to 39 years old. P(39 to 39) =________

Determine the probability that a randomly selected multiple birth for women 15-54 years old involved a mother who was not 30 to 39 years old. P (not 30 to 39) = _______

Determine the probability that a randomly selected multiple birth women 15-54 years old involved a mother who as less than 45 years old. P(less than 45)=_______

Determine the probability that a randomly selected multiple birth for women 15-54 years old involved a mother who as at least 20 years old. P(at least 20)=________

Age

Number of Multiple Births

15-19

91

20-24

513

25-29

1630

30-34

2839

35-39

1841

40-44

379

45-54

118

0 0
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Answer #1
Concepts and reason

Probability: The ratio of the number of favorable outcomes to certain event and total number of possible outcomes is called as the probability of an event.

Complementary event: If an event A is defined, then complement of event A is not occurring of event A.

Fundamentals

The probability of an event is defined as,

Probability=NumberoffavorableoutcomesforaneventTotalnumberofoutcomes{\rm{Probability}} = \frac{{{\rm{Number}}\,{\rm{of}}\,{\rm{favorable}}\,{\rm{outcomes}}\,{\rm{for}}\,{\rm{an}}\,{\rm{event}}}}{{{\rm{Total}}\,{\rm{number}}\,{\rm{of}}\,{\rm{outcomes}}}}

Define event A, some of the probability formulas are,

The complementary of event A is, P(A)=1P(A)P\left( {A'} \right) = 1 - P\left( A \right)

P(lessthanA)=1P(greaterthanA)P\left( {{\rm{less}}\,{\rm{than}}\,A} \right) = 1 - P\left( {{\rm{greater}}\,{\rm{than}}\,A} \right)

P(atleastA)=1P(atmostA)P\left( {{\rm{atleast}}\,A} \right) = 1 - P\left( {{\rm{atmost}}\,A} \right)

The probability of randomly selected multiple birth for women 15-54 years old involved a mother of age 30-39 is obtained as shown below:

From the information it is clear that for the age group 30-34 the number of multiple birth is 2839 and for the age group 35-39 the number of multiple birth is 1841.

The number of favorable outcome is 2,839+1,841=4,6802,839 + 1,841 = 4,680 and the total number of outcomes is 91+513+1,630+2,839+1,841+379+118=7,41191 + 513 + 1,630 + 2,839 + 1,841 + 379 + 118 = 7,411.

The required probability is,

Probability=NumberoffavorableoutcomesforaneventTotalnumberofoutcomes=4,6807,411=0.631\begin{array}{c}\\{\rm{Probability}} = \frac{{{\rm{Number}}\,{\rm{of}}\,{\rm{favorable}}\,{\rm{outcomes}}\,{\rm{for}}\,{\rm{an}}\,{\rm{event}}}}{{{\rm{Total}}\,{\rm{number}}\,{\rm{of}}\,{\rm{outcomes}}}}\\\\ = \frac{{4,680}}{{7,411}}\\\\ = 0.631\\\end{array}

The probability of randomly selected multiple birth for women 15-54 years old does not involved a mother of age 30-39 is,

From the information it is clear that for the age group 30-34 the number of multiple birth is 2839 and for the age group 35-39 the number of multiple birth is 1841.

The number of favorable outcome is 2,839+1,841=4,6802,839 + 1,841 = 4,680 and the total number of outcomes is 91+513+1,630+2,839+1,841+379+118=7,41191 + 513 + 1,630 + 2,839 + 1,841 + 379 + 118 = 7,411.

The required probability is,

Probability=1NumberoffavorableoutcomesforaneventTotalnumberofoutcomes=14,6807,411=10.631=0.369\begin{array}{c}\\{\rm{Probability}} = 1 - \frac{{{\rm{Number}}\,{\rm{of}}\,{\rm{favorable}}\,{\rm{outcomes}}\,{\rm{for}}\,{\rm{an}}\,{\rm{event}}}}{{{\rm{Total}}\,{\rm{number}}\,{\rm{of}}\,{\rm{outcomes}}}}\\\\ = 1 - \frac{{4,680}}{{7,411}}\\\\ = 1 - 0.631\\\\ = 0.369\\\end{array}

The probability of randomly selected multiple birth for women 15-54 years old involved a mother of age less than 45 is,

From the information it is clear that for the age group 45-49 the number of multiple births is 118.

The total number of outcomes is 91+513+1,630+2,839+1,841+379+118=7,41191 + 513 + 1,630 + 2,839 + 1,841 + 379 + 118 = 7,411.

The required probability is,

P(lessthan45)=1P(greaterthan45)=11187,411=10.0159=0.984\begin{array}{c}\\P\left( {{\rm{less}}\,{\rm{than}}\,45} \right) = 1 - P\left( {{\rm{greater}}\,{\rm{than}}\,\,45} \right)\\\\ = 1 - \frac{{118}}{{7,411}}\\\\ = 1 - 0.0159\\\\ = 0.984\\\end{array}

The probability of randomly selected multiple birth for women 15-54 years old involved a mother of age at least 20 is,

From the information it is clear that for the age group 15-19 the number of multiple births is 91.

The total number of outcomes is 91+513+1,630+2,839+1,841+379+118=7,41191 + 513 + 1,630 + 2,839 + 1,841 + 379 + 118 = 7,411.

The required probability is,

P(atleast20)=1P(atmost20)=1917,411=10.0123=0.988\begin{array}{c}\\P\left( {{\rm{at}}\,{\rm{least}}\,20} \right) = 1 - P\left( {{\rm{atmost}}\,20} \right)\\\\ = 1 - \frac{{91}}{{7,411}}\\\\ = 1 - 0.0123\\\\ = 0.988\\\end{array}

Ans:

Thus, the probability that the randomly selected multiple birth for women of age 15-54 involved a mother of age group 30-39 is 0.631.

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