The president of Doerman Distributors, Inc., believes that 28% of the firm's orders come from first-time customers. A simple random sample of 100 orders will be used to estimate the proportion of first-time customers. Use z-table.
What is the probability that the sample proportion will be between 0.16 and 0.40 (to 4 decimals)?
What is the probability that the sample proportion will be between 0.21 and 0.35 (to 4 decimals)?
1)
for normal distribution z score =(p̂-p)/σp | |
here population proportion= p= | 0.280 |
sample size =n= | 100 |
std error of proportion=σp=√(p*(1-p)/n)= | 0.0449 |
probability = | P(0.16<X<0.4) | = | P(-2.67<Z<2.67)= | 0.9962-0.0038= | 0.9924 |
2)
probability = | P(0.21<X<0.35) | = | P(-1.56<Z<1.56)= | 0.9406-0.0594= | 0.8812 |
The president of Doerman Distributors, Inc., believes that 28% of the firm's orders come from first-time...
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