Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed. a. Find the probability that at least 150 favor a charter school. b. Find the probability that at most 160 favor a charter school. c. Find the probability that more than 155 favor a charter school.
Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed. a. Find the probability that at least 150 favor a charter school. b. Find the probability that at most 160 favor a charter school. c. Find the probability that more than 155 favor a charter school.
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Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed.

Transcribed Image Text:Example 7.12
Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the population favor a
charter school for grades K through 5. A simple random sample of 300 is surveyed.
a. Find the probability that at least 150 favor a charter school.
b. Find the probability that at most 160 favor a charter school.
c. Find the probability that more than 155 favor a charter school.
d. Find the probability that fewer than 147 favor a charter school.
e. Find the probability that exactly 175 favor a charter school.
Let X = the number that favor a charter school for grades K trough 5. X ~ B(n, p) where n = 300 and p = 0.53.
Since np > 5 and nq > 5, use the normal approximation to the binomial. The formulas for the mean and standard
deviation are µ = np and ơ =
ynpq. The mean is 159 and the standard deviation is 8.6447. The random variable
for the normal distribution is Y. Y ~ N(159, 8.6447). See The Normal Distribution for help with calculator
instructions.
For part a, you include 150 so P(X > 150) has normal approximation P(Y > 149.5) = 0.8641.
normalcdf(149.5,10^99,159,8.6447) = 0.8641.
For part b, you include 160 so P(X < 160) has normal appraximation P(Y < 160.5) = 0.5689.
normalcdf(0,160.5,159,8.6447) = 0.5689
For part c, you exclude 155 so P(X > 155) has normal approximation P(y > 155.5) = 0.6572.
normalcdf(155.5,10^99,159,8.6447) = 0.6572.
For part d, you exclude 147 so P(X < 147) has normal approximation P(Y < 146.5) = 0.0741.
normalcdf(0,146.5,159,8.6447) = 0.0741
For part e,P(X = 175) has normal approximation P(174.5 < Y< 175.5) = 0.0083.
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