Question

en Find the indicated probability using the standard normal distribution Plz< -1.28 or z> 128) Click here to view page 1 of the standard normal table. Click here to view page 2 of the standard normal table P(z< - 1.28 or z>1.28)=(Round to four decimal places as needed.)

^these two are together

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^these two are together

^these two are together

Answer 1

ałt & tnt, & solution P[24-1.2008 271,20] P[ 24-1.201 + P(27) 201 2 P1221.28 Z 2X 1003 2006 ons -1.28 nzio, ła 3lil, 8= 713 951. Confidence interval for le Jn 1.28 Next Ques = Paris Hi-t, 10-1 los . 211098 31.1+ 7,3 x 2.1098 = 3l11t 3,6302 = [ 27.4699, 34.7302) Rounded value upto 1-decimal L27,5, 34,7) ms #margin oferow Rounded value Upto I decimal E = 3.6 oms Interpretation with 95%. Confidence It can be said that the population mean Connmute time is between the bound as the confidence Interval.

Ques of a 2.03 then Right tailed 2-critical z = 2.03 = 1.8808 Rounded value opto 2 2-decimal 12,03= Any h Option Rejection Region 1s/2> 1.08 -on 03=1.80 Ion In graph option VILLA 2 auus Of divo3,02 394 n = 6100, n=43 Hall a Alternate hypothesis coreco nool-Goo o Hit l 2600 test - statistics (2) at-uno (ارہ ) ) of Jn 6100-6000 394/563 2-stat Icox543 -1.6643 394 Rounded value of 12-stat 1. 66 6 # critical value for two tailed test option critical value £ 2017 2-stat az-Critical so option fail to Reject no s'At the so there is not enough evidence to 3% Level Support telect the claim 203 201701 t conclusion

2 (Gus) Шz съое Лами PL x737) 1 - 3 > 37 ) Pz 32 - se - se) (2) -), sə1] 0.5tProca 0222.05 17 — О, 460 ч 2.968 u Ques inean height (MZ- 6714,0 = 3 © Pixies) - Рf a < 6s P2K 6S- 62. 41 = P2 L - 0 0 ) 3 - 2 119 — zzo 6 P[LS < x < 0) fest 23 PS-62.4 3 2 < 20 – 624 3 3 PC - 0.0 ( 2 8 667) — рос 2 с. 8+) Рfoc2484417 2001 + 3669 5950 mg 2 со . 8 66 9 - 1 .

ou U240xz.og, o23.08, X = 38,3, n260 P{ 27 70] = { my 770-MJ = P{27 70-67.1 =P[27, 8667] = 1931 ms d ③ - The events in part @ @ are unusual because its probabilities is of less than, os Quis given confidence interval for p L. 664,692 + E 664 1 itt = .692 Adding Oh we get 664 t, 692 = 1.356 Tha 11356 678 - $ 26. Again 1 subtracting from the get 2€ 1692- 664 = 1028 olu 2 = Nulla Alternative hypothesis option (B Ho! Me Go Us Hil840 Standard-2-stat = | r-uno 138.3-40 3108 1 66 2-stat -1.70x56. 3108 4.2754

Rounded value of/2- 2-stat & 4.28 A critical value = £1,70 4 Conclusion: tytuto fail to Bog Rejectho because 2-stat is greater a than 2 - Critical Reject no, At gy. significance level, there is enough evidence to Reject the claim. # The critical value is 1 2212