In: Statistics and Probability
1. It is known that 15.2% of the population is left-handed. Select a sample of 60 individuals? a) Describe the sampling distribution of the sample proportion. Be sure to calculate the mean (expected value) and the standard deviation of the sample proportion. b) What is the probability that at least 20% of individuals in this sample will be left-handed? c) What is the probability that at most 35% of individuals in this sample will be left-handed?
Solution :
a) If p is a population proportion and a sample of size n is taken from this population and if np ≥ 5, nq ≥ 5, then sampling distribution of sample proportions (p̂) of all the samples of size n will be approximately normal with mean p and standard deviation .
i.e.
(Where, q = 1 - p)
Population proportion of left-handed (p) = 15.2% = 0.152
Sample size (n) = 60
q = 1 - 0.152 = 0.848
np = 60 × 0.152 = 9.12 which is greater than 5.
no = 60 × 0.848 = 50.88 which is greater than 5.
Hence, sampling distribution of sample proportion will be approximately normal with mean p and standard deviation .
The mean of the sample proportions is given as follows:
The mean of the sample proportions is 0.152.
The standard deviation of the sample proportions is given as follows :
The standard deviation of the sample proportions is 0.04635.
Hence, sampling distribution of sample proportion would be approximately normal with mean 0.152 and standard deviation 0.04635.
b) We have to obtain Pr(p̂ ≥ 0.20).
We have, p = 0.152, q = 0.848 and n = 60
We know that if ~ N(p, pq/n) then,
Using "pnorm" function of R we get, Pr(Z ≥ 1.0356) = 0.1502
Hence, the probability that at least 20% of individuals in this sample will be left-handed is 0.1502.
c) We have to obtain Pr(p̂ ≤ 0.35).
We have, p = 0.152, q = 0.848 and n = 60
We know that if ~ N(p, pq/n) then,
Using "pnorm" function of R we get, Pr(Z ≤ 4.2719) = 1.0000
Hence, the probability that at most 35% of individuals in this sample will be left-handed is 1.0000.
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