Question

Ex: 24,26,28,30,32,34,36

abis standard deviation of I. In each case, draw a graph, then find the probablity o of o aven bone density tesf scores. I using technology instead of Table A-2, round ans rmal Distribution. In Exercises 17-36, assume that a randomly selected a bone density test. Those test scores are normally distributed with a mea rdecimal places 17. Less than 1.23 18. Less than -1.96 20. Less than 2.56 22. Greater than 0.18 24. Greater than -3.05 26. Between 1.50 and 2.50 28. Between-2.75 and-0.75 30. Between-3.00 and 3.00 32. Between-4.27 and 2.34 34. Greater than -3.75 19. Less than 1.28 21. Greater than 0.25 23. Greater than-2.00 25. Between 2.00 and 3.00 27. Between and -2.55 and -2.00 29. Between -2.00 and 2.00 31. Between -1.00 and 5.00 33. Less than 4.55 35. Greater than 0 36. Less than 0

Answer 1

Ans. Z be bone density test scores, has normal distribution with mean = 0 and SD = 1

(24). P(Z>3.05) = 0.5 - P(0 < Z < 3.05)

= 0.5 - 0.4989 = 0.0011

(26).(1.50 < Z < 2.50) = P(0 < Z < 2.50) - P(0 < Z < 1.50)

= 0.4938 - 0.4332 = 0.0606

(28). P(-2.75 < Z < - 0.75) = P(0.75 < Z < 2.75), since Z is symmetrical distribution about Z = 0

=P(0 < Z < 2.75) - P( 0. < Z < 0.75)

= 0.4970 - 0.2734 = 0.2236

(30). P(-3.0 < Z < 3.0) = 2*P(0 < Z < 3.0)

=2*0.4987 = 0.9974

(32).P(-4.27 < Z < 2.34) = P(0 < Z < 4.27) + P(0 < Z < 2.34)

= 0.5000 + 0.4904 = 0.9904

(34). P(Z > - 3.75) = 0.5 + P(0 < Z < 3.75)

= 0.5 + 0.4999 = 0.9999

(36).P(Z > 0) = 0.5000