MTH 245.19PR NAME___________________ MTH 245.LO19 XXXXXXXXXXTEST 3 SHOW ALL WORK NO CREDIT WILL BE GIVEN FOR ANY PROBLEM FOR WHICH NO WORK IS SHOWN-EVEN IF THE ANSWER IS CORRECT 1 1. Use the standard...

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MTH 245.19PR NAME___________________ MTH 245.LO19 TEST 3 SHOW ALL WORK NO CREDIT WILL BE GIVEN FOR ANY PROBLEM FOR WHICH NO WORK IS SHOWN-EVEN IF THE ANSWER IS CORRECT 1 1. Use the standard normal table to find each of the following. Be sure to include a sketch of the curve as part of the answer. a.  P 1.73 Z 2.52   Ans 1a___________ b.  P Z 2.18 Ans 1b___________ c.  P 1.46 Z 6.09  Ans 1c___________ d.  P 2.54 Z .63    Ans 1d___________ e. find 0z such that  0P Z z  .7486 Ans 1e___________ MTH 245.19PR NAME___________________ MTH 245.LO19 TEST 3 SHOW ALL WORK NO CREDIT WILL BE GIVEN FOR ANY PROBLEM FOR WHICH NO WORK IS SHOWN-EVEN IF THE ANSWER IS CORRECT 2 2. On a golf driving range, driving distances are normally distributed with a mean of 208 yards and a standard deviation of 14.2 yards. What is the probability of driving at least 248.47 yards? Ans 2____________ 3. Random samples of size 28 are drawn from a population with a mean of 326.8 and a variance of 17.3. If the sampling distribution of X is constructed, what would be the values of X  , 2 X  , and X  ? Ans 3 X  ______ 2 X   ______ X   ______ 4. a. How would you interpret a confidence level of 86% when n  400 and you are estimating ? b. If you construct 7000 86% C.I.’s, how many of those intervals would contain the true value of ? Ans 4b___________ MTH 245.19PR NAME___________________ MTH 245.LO19 TEST 3 SHOW ALL WORK NO CREDIT WILL BE GIVEN FOR ANY PROBLEM FOR WHICH NO WORK IS SHOWN-EVEN IF THE ANSWER IS CORRECT 3 5. A continuous r.v. X can assume any value (with equal likelihood) over the interval from 80 to 140. a. Draw the density curve that describes the distribution of this r.v. and calculate  f x . b. Find  P 93.5 X 119.3  Ans 5b___________ c. What value would you expect X to assume? Ans 5c___________ 6. X is binomial with n  800 and p  .32. Use the standard normal distribution to approximate: a.  P 225 X 271  Ans 6a___________ b.  P X 290 Ans 6b___________ c. What allows us to use the standard normal distribution to approximate binomial probabilities when n is large? Ans 6c___________ MTH 245.19PR NAME___________________ MTH 245.LO19 TEST 3 SHOW ALL WORK NO CREDIT WILL BE GIVEN FOR ANY PROBLEM FOR WHICH NO WORK IS SHOWN-EVEN IF THE ANSWER IS CORRECT 4 7. A recent poll of 4000 high school teachers shows that 1360 of them favor block scheduling. a. Find a point estimate for the proportion of all high school teachers that favor block scheduling. Ans 7a_____________ b. Estimate the true proportion of all high school teachers that favor block scheduling using an 92% C.I. Ans 7b_____________ 8. In a random sample of 18 fifth graders, the mean weight is 62.8 pounds and the standard deviation is 9.4 pounds. Estimate the mean weight of all fifth graders using a 99% C.I. Ans 8_______________ 9. From the time of diagnosis, a random sample of 100 liver cancer patients has a mean life expectancy of 6 months with a standard deviation of .35 months. Estimate the mean life expectancy for all liver cancer patients using a 97% C.I. Ans 9________________
Answered Same DayNov 08, 2021

Answer To: MTH 245.19PR NAME___________________ MTH 245.LO19 XXXXXXXXXXTEST 3 SHOW ALL WORK NO CREDIT WILL BE...

Rajeswari answered on Nov 08 2021
110 Votes
Normal distribution assignment
a)
Area = 0.4941+0.4582=0.9523
b)
P(Z<2.18) = 0.5+0.4854 =0.985
4
c) P(1.46d) P(-2.54e) P(ZZ0=0.67
2. On a golf driving range, driving distances are normally distributed with a mean of 208 yards and a standard deviation of 14.2 yards. What is the probability of driving at least 248.47 yards?
Reqd prob = P(X>248.47) (x- driving distance)
= P(
3. Random samples of size 28 are drawn from a population with a mean of 326.8 and a variance of 17.3. If the sampling distribution of X is constructed, what would be the values of X , 2 X , and X ?
N = 28: Mean =326.8 and var = 17.3
By central limit theorem sample mean = mu = 326.8
Sample variance of mean...
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