MATH 3339 1 CERTIFICATION SCRIPT 2026
QUESTIONS WITH SOLUTIONS GRADED A+
◍ x bar= 26.6n = 81sigma = 6x bar +/- z(sigma/sqrt(n))>
26.5+c(-1,1)*qnorm(1.98/2)*(6/sqrt(81))[1] 24.9491 28.0509.
Answer: The gas mileage for a certain model of car is known to have a
standard deviation of 6 mi/gallon. A simple random sample of 81 cars of
this model is chosen and found to have a mean gas mileage of 26.5
mi/gallon. Construct a 98% confidence interval for the mean gas mileage for
this car model.
◍ a) The sample mean is an unbiased estimate of the population mean..
Answer: Which of the following statements are true?a) The sample mean is
an unbiased estimate of the population mean.b) The z-critical value is an
estimate of the population mean.c) The t-critical value is an estimate of the
population standard deviation.d) The sample standard deviation is an
unbiased estimate of the population standard deviation.
◍ n < 30, therefore t-test intervaln =22x bar = 5.6sd = 0.1c = 0.90>
5.6+c(-1,1)*qt(1.9/2,21)*(.1/sqrt(22))[1] 5.563314 5.636686.
Answer: An SRS of 22 students at UH gave an average height of 5.6 feet
and a standard deviation of .1 feet. Construct a 90% confidence interval for
the mean height of students at UH.
◍ Simplify Radical -1/25.
Answer: i/5
◍ the area under the graph..
Answer: pdf focuses more on..
◍ Algebraic Solving.
Answer: Algebraic Solving
, ◍ decrease the variance.
Answer: What will reduce the width of a confidence interval?
◍ P(X < 79)sigma = sqrt(p(1-p)/n) sqrt((0.83*(1-0.83))/144)[1]
0.03130273Now we have μ, n, and x, so X~norm>
pnorm(0.79,0.83,0.0313)[1] 0.1006326.
Answer: In a large population, 83% of the households have cable tv. A
simple random sample of 144 households is to be contacted and the sample
proportion computed. What is the probability that the sampling distribution
of sample porportions is less than 79%?
◍ n = 36sigma = 0.1 lb = 1.6 ozμ = 1 lb = 16 ozP(X bar < 15 oz)>
pnorm(15,16,1.6/sqrt(36))[1] 8.841729e-05 or 0.0001.
Answer: Lloyd's Cereal company packages cereal in 1 pound boxes (16
ounces). A sample of 36 boxes is selected at random from the production
line every hour, and if the average weight is less than 15 ounces, the
machine is adjusted to increase the amount of cereal dispensed. If the mean
for 1 hour is 1 pound and the standard deviation is 0.1 pound, what is the
probability that the amount dispensed per box will have to be increased?
◍ Simplify i^47.
Answer: -i
◍ Convert 9/13 into percentage.
Answer: 9/13 0.6923 69%
◍ right tailed , so 1-c = 1-0.54 = 0.46 1 - qnorm(0.46) = -0.10.
Answer: Find a value of c so that P(Z ≤ c) = 0.71.
◍ decreases.
Answer: As the length of the confidence interval for the population mean
decreases, the degree of confidence in the interval's actually containing the
population mean
◍ Solve for x in x^3 + 27 = 0.
Answer: x^3 + 27 = 0x^3 = -27x^3(1/3) = -27^1/3x=-3
QUESTIONS WITH SOLUTIONS GRADED A+
◍ x bar= 26.6n = 81sigma = 6x bar +/- z(sigma/sqrt(n))>
26.5+c(-1,1)*qnorm(1.98/2)*(6/sqrt(81))[1] 24.9491 28.0509.
Answer: The gas mileage for a certain model of car is known to have a
standard deviation of 6 mi/gallon. A simple random sample of 81 cars of
this model is chosen and found to have a mean gas mileage of 26.5
mi/gallon. Construct a 98% confidence interval for the mean gas mileage for
this car model.
◍ a) The sample mean is an unbiased estimate of the population mean..
Answer: Which of the following statements are true?a) The sample mean is
an unbiased estimate of the population mean.b) The z-critical value is an
estimate of the population mean.c) The t-critical value is an estimate of the
population standard deviation.d) The sample standard deviation is an
unbiased estimate of the population standard deviation.
◍ n < 30, therefore t-test intervaln =22x bar = 5.6sd = 0.1c = 0.90>
5.6+c(-1,1)*qt(1.9/2,21)*(.1/sqrt(22))[1] 5.563314 5.636686.
Answer: An SRS of 22 students at UH gave an average height of 5.6 feet
and a standard deviation of .1 feet. Construct a 90% confidence interval for
the mean height of students at UH.
◍ Simplify Radical -1/25.
Answer: i/5
◍ the area under the graph..
Answer: pdf focuses more on..
◍ Algebraic Solving.
Answer: Algebraic Solving
, ◍ decrease the variance.
Answer: What will reduce the width of a confidence interval?
◍ P(X < 79)sigma = sqrt(p(1-p)/n) sqrt((0.83*(1-0.83))/144)[1]
0.03130273Now we have μ, n, and x, so X~norm>
pnorm(0.79,0.83,0.0313)[1] 0.1006326.
Answer: In a large population, 83% of the households have cable tv. A
simple random sample of 144 households is to be contacted and the sample
proportion computed. What is the probability that the sampling distribution
of sample porportions is less than 79%?
◍ n = 36sigma = 0.1 lb = 1.6 ozμ = 1 lb = 16 ozP(X bar < 15 oz)>
pnorm(15,16,1.6/sqrt(36))[1] 8.841729e-05 or 0.0001.
Answer: Lloyd's Cereal company packages cereal in 1 pound boxes (16
ounces). A sample of 36 boxes is selected at random from the production
line every hour, and if the average weight is less than 15 ounces, the
machine is adjusted to increase the amount of cereal dispensed. If the mean
for 1 hour is 1 pound and the standard deviation is 0.1 pound, what is the
probability that the amount dispensed per box will have to be increased?
◍ Simplify i^47.
Answer: -i
◍ Convert 9/13 into percentage.
Answer: 9/13 0.6923 69%
◍ right tailed , so 1-c = 1-0.54 = 0.46 1 - qnorm(0.46) = -0.10.
Answer: Find a value of c so that P(Z ≤ c) = 0.71.
◍ decreases.
Answer: As the length of the confidence interval for the population mean
decreases, the degree of confidence in the interval's actually containing the
population mean
◍ Solve for x in x^3 + 27 = 0.
Answer: x^3 + 27 = 0x^3 = -27x^3(1/3) = -27^1/3x=-3
