Random samples of size n = 80 were selected from a binomial population with p = 0.6. Use the normal distribution to approximate the following probabilities. (Round your answers to four decimal places.)
(a)
P(p̂ ≤ 0.63) =
(b)
P(0.57 ≤ p̂ ≤ 0.63) =
To approximate the probabilities using the normal distribution, we need to use the sample mean, p̂, as an estimator for the population proportion, p. The sample mean is given as p̂ = X/n, where X is the number of successes in the sample and n is the sample size.
(a) P(p̂ ≤ 0.63):
To calculate this probability, we need to calculate the Z-score corresponding to the value of p̂ = 0.63 using the formula: Z = (p̂ - p)/sqrt(p(1-p)/n), where p̂ = 0.63, p = 0.6, and n = 80.
Z = (0.63 - 0.6)/sqrt(0.6(1-0.6)/80)
= 0.03/sqrt(0.24/80)
= 0.03/0.049
Next, we can find the probability using the Z-table (a table that gives the area under the standard normal curve). We need to find the probability of Z ≤ calculated Z-score.
Using the Z-table or a calculator, we find that the probability P(Z ≤ 0.6122) is approximately 0.7293.
Hence, P(p̂ ≤ 0.63) ≈ 0.7293 (rounded to four decimal places).
(b) P(0.57 ≤ p̂ ≤ 0.63):
To calculate this probability, we need to calculate the Z-scores for both p̂ values.
For p̂ = 0.57:
Z1 = (0.57 - 0.6)/sqrt(0.6(1-0.6)/80)
= -0.03/0.049
For p̂ = 0.63:
Z2 = (0.63 - 0.6)/sqrt(0.6(1-0.6)/80)
= 0.03/0.049
Next, we need to find the probability P(Z1 ≤ Z ≤ Z2).
Using the Z-table or a calculator, we find that the probability P(-0.6122 ≤ Z ≤ 0.6122) is approximately 0.4767.
Hence, P(0.57 ≤ p̂ ≤ 0.63) ≈ 0.4767 (rounded to four decimal places).