By using z table we get P( z < z cri ) =0.0580
P( z < -1.57 ) = 0.0580 ( Using table )
z critical = -1.57
8) d) -1.57
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8) Suppose that p(z sz) = .0580 Using the z-table determine the closest value for zc...
Let z have a normal standard distribution. Determine the value of Zc. P(0 < Z < Zc)=0.4573
b. P(-1.69 SZ SZ) = 0.9545 C. P(Z > 20) = 0.0048 d. P(Z, SZS 2.00) = 0.1359 e. P(Z > Z) = 0.9713 f. PC-Z, SZSZ) = 0.2358
Using the normal table or software, find the value of z that makes the following probabilities true. You might find it helpful to draw a picture to check your answers. (a) P(Z <z) = 0.40 (b) P(Z = z) = 0.50 (c) P(-zsZ sz) = 0.50 (d) P(|Z| > Z) = 0.01 (e) P(|Z| <z) = 0.90 (a) z= (Round to four decimal places as needed.)
In each case, determine the value of the constant c that makes the probability statement correct. (Round your answers to two decimal places.) Ф(c)-0.9838 (a) (b) P(O SZ sc) 0.2967 (c) PC s Z)0.1230 (d) P(-c c)-0.6424 Z (e) P(c s IZ1)-0.0160
In each case, determine the value of the constant c that makes the probability statement correct. (Round your answers to two decimal places.) Ф(c)-0.9838 (a) (b) P(O SZ sc) 0.2967 (c) PC s Z)0.1230 (d) P(-c c)-0.6424 Z...
3. Find the value of z such that the following are satisfied: a. P(Z < z) = 0.85 b. P(Z <= z) = 0.5 C. P(Z > z) = 0.85 d. P(-1.24 < Z < z) = 0.85
Hypothesis Testing, P-values: For the given alternate hypothesis (H1) and test statistic (z), determine the P-value of the test statistic. Use the answer found in the z-table or round to 4 decimal places. (a) H1: p < 0.75 The test statistic is z = −2.83. P-value = (b) H1: p > 0.85 The test statistic is z = 1.84. P-value = (c) H1: p ≠ 0.15 The test statistic is z = 1.85. P-value =
3. P(z<zc)=0.95. Find ze (a) 1.28 (b) 1.645 (c) 1.96 (d) -1.645
Consider using a z test to test Ho: p = 0.3. Determine the p-value in each of the following situations. (Round your answers to four decimal places.) (a) Ha:p > 0.3, z = 1.46 (b) H:P < 0.3, z = -2.77 (c) H.: p0.3, z = -2.77 (d) H.:P <0.3, z = 0.23 You may need to use the appropriate table in the Appendix of Tables to answer this question.
We can now use the Standard Normal Distribution Table to find the probability P(-0.25 sz s 1). 0.05 0.06 0.07 0.08 0.09 -0.2 0.4013 0.3974 0.3936 0.3897 0.3859 0.00 0.01 0.02 0.03 0.04 Using these 1.0 0.8413 0.8438 0.8461 0.8485 0.8531 The table entry for z = -0.25 is 0.00 and the table entry for z = 1 is values to calculate the probability gives the following result. PC-0.25 sz s 1) P(Z < 1) - P(Z 5 -0.25) 10....
7. Hint for c and d: given P(X S x) a percentage, we have P(Z Sz)the percentage. Then find the corresponding value for Z, and use the Inverse Transformation Let X be normally distributed with mean 120 and standard deviation σ 20. a. Find P(X3 86). b. Find P(80 <X3100). c. Find x such that P(Xx) 0.40. d. Find x such that P(X> x) 0.90.