a) 37/75
b) If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
(37/75)^2 * (38/75)^3 = ?
C) do similar process as b.
The correct size of a nickel is 21.21 millimeters. Based on that, the data can be summarized into the following table:
Too Small-Too Large- Total
Low Income 11 29 40
High Income 26 9 35
Total 37 38 75
Based on this data: (give your answers to parts a-c as fractions, or decimals to at least 3 decimal places. Give your to part d as a whole number.)
a) The proportion of all children that drew the nickel too small is:
Assume that this proportion is true for ALL children (e.g. that this proportion applies to any group of children), and that the remainder of the questions in this section apply to selections from the population of ALL children.
b) If 5 children are chosen, the probability that exactly 2 would draw the nickel too small is:
c) If 5 children are chosen at random, the probability that at least one would draw the nickel too small is:
2 answers
"
b) If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
(37/75)^2 * (38/75)^3 = ?
"
The reason this doesn't work is that there are different permutations of ways the children can be arranged (there's 10, Let me explain). If you do what PsyDAG said then you find the probability that the FIRST two kids draw too small and the LAST three are two big, but that is not the same are any old "two kids". To get the number of permutations use the P-Rule (n!/(n!p*n!q*n!k where the p q and k are subscript to the respective n!). In this case:
5!/(2! * 3!) = 10
And then multiply the answer from PsyDAG's Equation by that 10.
Or you can use a Binomial Function from Excel or TI Calc (found in dist):
binomcdf(5, 38/75, 2)
=BINOM.CDF(n: 5, p: 38/75, x: 2)
Hope this helps!
b) If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
(37/75)^2 * (38/75)^3 = ?
"
The reason this doesn't work is that there are different permutations of ways the children can be arranged (there's 10, Let me explain). If you do what PsyDAG said then you find the probability that the FIRST two kids draw too small and the LAST three are two big, but that is not the same are any old "two kids". To get the number of permutations use the P-Rule (n!/(n!p*n!q*n!k where the p q and k are subscript to the respective n!). In this case:
5!/(2! * 3!) = 10
And then multiply the answer from PsyDAG's Equation by that 10.
Or you can use a Binomial Function from Excel or TI Calc (found in dist):
binomcdf(5, 38/75, 2)
=BINOM.CDF(n: 5, p: 38/75, x: 2)
Hope this helps!
1 answer