21. A 95% confidence interval estinate can be interpreted to mean that
(A) Ifall possible samples of size N are taken and confidence interval estimates are developed, 95% of thems would include tibe true populat tion tean somewhere within their interval.
(B) We have 95% confidence that we have selected a sample whose interval does include the he population mean.
(C) Both above.
(D) None of the above.

1. Let = 3600.7 and S2 = 14655 be the sample mean and variance of the weights of nineteen neonates in grams. Suppose that there was a typewriting error in the cighth observation, the accurate weight of eighth neonate, 4320 was recorded as 4230. After we correct it, what are the accurate sample mean and variance (A) Sample mean: 3605.44, Variance: 20274.32 (B) Sample mea an: 3605.44, Varianc ce: 21175.7 (C) Sample mean: 3609.23, Variance: 22437.62 (D) Sample mean: 3609.23, Variance: 23416.3

2. Let A and B be the events such that P( A ) = 0.4, P( B ) = 0.6, and P(AC ∩ BC)=, Then P(AC∪BC) =?,P(AC ∪ BC) =? (A) (B) (C) (D)

3. Suppose that a firn's sales were S2,500,000 four year ago, and sales have grown annually by 1 5%, 15%, -10%, 5% since that time. What was the mean growth rate in sales over the past four years? (A) 6% (B) 17% (C)25% (D) 33%

4. An insurance company believes that people can be divided into two classes - those are accident-prone and those are not. Their statistics show that an accident-prone person will have an accident at some time within a fixed I-year period with probability 0.5, whereas this probability decreases to 0.4 for a non-accident-prone person. If we assume that 40% of the pcpulation is accident-prone, what is the probability that a new policy holder will have an accident within a year of purchasing this policy? (A)0.44 (B) 0.45 (C) 0.46 (D) 0.47

5. Given the probabilities of some of events as follows: P(C|B) =0.95, P(C|BC)=0.3, P(AIB∩ C) = 0.2, P(A|BC∩ C) = 0.05, P(A|B ∩ C) = 0.35, P(AIBC ∩CC)= 0.2.Compute P(A|B). (A)0.2021 (B)0.2075 (C) 0.2098 (D)0.2132

6. A Bayesian approach can be used to revise probubilitis that a prospect field will produce oil. In one case, gcological assessment indicates a 25% chance that the field will produce oil. Furthes, tlere is an 80% chance that particular well will strike oil given that oil is preseat in the prospect filed, Suppose that one well is drilled on the field and it comes up dry. What is the probability that tbe prospect filed will produce oil? (A) 0.0333 (B)0.0425(C)0.0625 (D) 0.0675

7. Suppose we took a pair of randor: samples (X1,X2), with replacement frot: pogulation f(X): Find E() and P(X＞1) (A) E() - 14, P(＞1)=0.50 (B) E () = 1.4, P( ＞1)=0.49 (C) E () =1.6. P(＞1)=0.69 (D) E ()= 1.6, P(＞ 1)=0.71

8. The Sutton police dcpartnent nust write, on average, 6 tickets a day to keep department revenues at budgeted levels. mean of 6.5 tickes per day. Interpret Suppose the number of tickets written per day follows a Poisson distribution with a t the value of the mean. (A) The mean has no interpretation. (B) The expecto ted pumber of tickets writt written would be 6.5 per day. (C) Halt of the days have less tharn 6.5 tickets writen and balf of the days have more than 6,5 tickets written. (D) The number of tickets that is written most often is 6.5 tickets per day.

9. A firm manufactures metal rods which must be rejected if they are not betwecn 6.025 and 6.475 cn in dianeter. The expect value of the diameter of the rods is 6.25 cmn, and the standard dev deviation of tbese diamet eter is 0.035 cm. What is the maximum probability that a rod will be rejected? (A) 0.0531 (B) 0.0444 (C) 0.0357 (D)0.0242

10. Consider tbe infinite population described by the probability distribution show here. Assume a rindom sumple of n = 80 measurenents is selected from the population, find the sampling distrioution of the sample mean  and calculate P (0.905＜＜1.095) (A) = 1. P(0.905 ＜ ＜ 1.905) =0,7330 (B)= 1, P(0.905＜ ＜1.905) = 0.6578 (C) = 1.2. P(0.905 ＜＜1.905)= 0.7920 (D)= 1.2, P(0.905＜＜ 1.905)=0.9505

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