MATH 1280 Self -Quiz Unit 7 (100%) – University of the People Grade 10.00 out of 10.00 ( 100%) Information text Recall that the population average of the heights in the file "pop1.csv" is μ = 170.035. Using simulation it can be shown that the probability of the sample average of the height falling within 2 centimeter of the population average is approximately equal to 0.925. From the simulations we also got that the standard deviation of the sample average is (approximately) equal to 1.122. In the next 3 questions you are asked to apply the Normal approximation to the distribution of the sample average using this information. The answer may be rounded up to 3 decimal places of the actual value: Question 1 Correct Question text Using the Norm al approximation, the probability that sample average of the heights falls within 2 centimeter of the population average is Answer:
0.9253374 Feedback: Using the Normal approximation, the computation of a probability associated with the random variable is conducted with the functions of the Normal distribution for the same expectation and standard deviation as the original distribution. The expectation is μ = 170.035. The standard deviation is σ = 1.122. The event corresponds to the interval [μ - 2, &mu + 2 ]. Therefore, the approximated probability is > mu <- 170.035 > sig <- 1.122 > pnorm(mu+2,mu,sig) - pnorm(mu -2,mu,sig) [1] 0.9253374 The correct answer is: 0.9253374 Question 2 Correct Question text Using the Normal approximation we get that the central region that contains 90% of the distribution of the sample average is of the form 170.035 ± z · 1.122. The value of z is Answer:
1.644854 Feedback: The structure of the central region that contains 90% of the Normal distribution is μ ± qnorm(0.95) · σ. Ho wever, μ = 170.035 and σ = 1.122. Therefore, z = qnorm(0.95) = 1.644854. The correct answer is: 1.644854 Question 3 Correct Question text Using the Normal approximation, the probability that sample average of the heights is less than 169 is Answer:
0.1781444 Feedback: Using the Normal approximation, the computation of a probability associated with the random variable is conducted with the functions of the Normal distribution for the same expectation and standard deviation as the original distribution. The expe ctation is μ = 170.035. The standard deviation is σ = 1.122. The event corresponds to the values less than 169. Therefore, the approximated probability is > mu <- 170.035 > sig <- 1.122 > pnorm(169,mu,sig) [1] 0.1781444 The correct answer is: 0.1781444 Que stion 4 Correct Question text
0.9253374 Feedback: Using the Normal approximation, the computation of a probability associated with the random variable is conducted with the functions of the Normal distribution for the same expectation and standard deviation as the original distribution. The expectation is μ = 170.035. The standard deviation is σ = 1.122. The event corresponds to the interval [μ - 2, &mu + 2 ]. Therefore, the approximated probability is > mu <- 170.035 > sig <- 1.122 > pnorm(mu+2,mu,sig) - pnorm(mu -2,mu,sig) [1] 0.9253374 The correct answer is: 0.9253374 Question 2 Correct Question text Using the Normal approximation we get that the central region that contains 90% of the distribution of the sample average is of the form 170.035 ± z · 1.122. The value of z is Answer:
1.644854 Feedback: The structure of the central region that contains 90% of the Normal distribution is μ ± qnorm(0.95) · σ. Ho wever, μ = 170.035 and σ = 1.122. Therefore, z = qnorm(0.95) = 1.644854. The correct answer is: 1.644854 Question 3 Correct Question text Using the Normal approximation, the probability that sample average of the heights is less than 169 is Answer:
0.1781444 Feedback: Using the Normal approximation, the computation of a probability associated with the random variable is conducted with the functions of the Normal distribution for the same expectation and standard deviation as the original distribution. The expe ctation is μ = 170.035. The standard deviation is σ = 1.122. The event corresponds to the values less than 169. Therefore, the approximated probability is > mu <- 170.035 > sig <- 1.122 > pnorm(169,mu,sig) [1] 0.1781444 The correct answer is: 0.1781444 Que stion 4 Correct Question text
