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:
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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 = ?
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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!
a) The proportion of all children that drew the nickel too small is given by the ratio of the number of children who drew the nickel too small to the total number of children. So, the proportion is:
(11 + 26) / 75 = 37 / 75 ≈ 0.493
b) To find the probability that exactly 2 children out of 5 draw the nickel too small, we need to calculate the probability of selecting 2 children who drew it small, and 3 children who did not.
Probability of selecting 2 children who drew it small:
(11/75) * (26/74) ≈ 0.052
Probability of selecting 3 children who did not draw it small:
(38/73) * (37/72) * (36/71) ≈ 0.280
Probability = (0.052) * (0.280) ≈ 0.014
c) To find the probability that at least one child out of 5 draws the nickel too small, we can calculate the probability of the complement event (none of the children drawing it too small) and subtract it from 1.
Probability of selecting 5 children who did not draw it small:
(38/75) * (37/74) * (36/73) * (35/72) * (34/71) ≈ 0.259
Probability = 1 - 0.259 ≈ 0.741
a) To find the proportion of all children that drew the nickel too small, we need to divide the number of children who drew the nickel too small by the total number of children.
The number of children who drew the nickel too small is given in the table as 11 for Low Income and 26 for High Income, totaling to 11 + 26 = 37.
The total number of children is given as 75 in the Total row of the table.
Therefore, the proportion of all children that drew the nickel too small is:
37 (number of children who drew the nickel too small) / 75 (total number of children) = 0.493 or 49.3% (to three decimal places).
So, the answer is 49.3% or 0.493.
b) The probability of choosing exactly 2 children who drew the nickel too small can be calculated using the binomial probability formula:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
where n is the total number of trials, k is the number of successful trials, and p is the probability of success.
In this case, n = 5 (as 5 children are chosen), k = 2 (as we want exactly 2 children to draw the nickel too small), and p = 0.493 (as calculated in part a).
Using the binomial probability formula:
P(X=2) = (5 choose 2) * (0.493)^2 * (1-0.493)^(5-2)
Calculating this will give us the probability.
(5 choose 2) = 5! / (2! * (5-2)!) = 10
(0.493)^2 = 0.243049
(1-0.493)^(5-2) = 0.270008
P(X=2) = 10 * 0.243049 * 0.270008 = 0.655 or 65.5% (to three decimal places).
So, the probability that exactly 2 children would draw the nickel too small is 65.5% or 0.655.
c) To find the probability that at least one child would draw the nickel too small, we need to calculate the probability of the complement event, i.e., the probability that none of the children would draw the nickel too small, and then subtract it from 1.
The probability that none of the children would draw the nickel too small can be calculated as:
P(X=0) = (5 choose 0) * (0.493)^0 * (1-0.493)^(5-0)
Calculating this will give us the probability.
(5 choose 0) = 1
(0.493)^0 = 1
(1-0.493)^(5-0) = 0.027557
P(X=0) = 1 * 1 * 0.027557 = 0.027557
The probability that at least one child would draw the nickel too small is:
P(at least one child) = 1 - P(none of the children) = 1 - 0.027557 = 0.972 or 97.2% (to three decimal places).
So, the probability that at least one child would draw the nickel too small is 97.2% or 0.972.
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.