1. Statistics Practice Quest 2 OBJ: Review standard normal distribution, confidence intervals and regression

2. 1. Determine the area under the standard normal curve that lies to the right of - 1.1. 4 th type of problem – look up on table.3643.5 z -3 -2 -1 0 1 2 3 -1.1 & add to.5.3643 +.5 =.8643

3. 2. Determine the area under the standard normal curve that lies to the left of z = - 2.75. 2 nd type of problem – look up on table.003.4970.5 z -3 -2 -1 0 1 2 3 -2.75 & subtract from.5.5 –.4970 =.003

4. 3. Determine the area under the standard normal curve that lies between -.38 and 1.2 3 rd type of problem – look up on table.1480.3849 z -3 -2 -1 0 1 2 3 -.38 1. 2 & add.1480 +.3849 =.5329

5. 4. Determine the area under the standard normal curve that lies to the left of z = -.72. 2 nd type of problem – look up on table.5 z -3 -2 -1 0 1 2 3 -2.75 & subtract from.5.5 –.2642 =.2358

6. 5. Determine the area under the standard normal curve that lies between.81 and 1.36 5th type of problem – look up on table.2910 z -3 -2 -1 0 1 2 3.81 1.36 & subtract.4131 –.2910 =.1221

7. Draw and shade area to find a value greater than 55. z = x – x s z = 55 – 50 5 z = 1 Given a normally distributed population with mean = 50 and standard deviation = 5. Each data item x can be converted to a standard value z by the formula: z = x – x s.3413 z - 3 -2 -1 0 1 2 3 1.4.5 –.3413.1587

8. Draw and shade area to find a value less than 42. z = x – x s z = 42 – 50 5 z = – 1.6 Given a normally distributed population with mean = 50 and standard deviation = 5. Each data item x can be converted to a standard value z by the formula: z = x – x s z - 3 -2 -1 0 1 2 3 – 1.6.5 –.4452 =.0548.

9. z = x – x s z = 48 – 50 5 z = -.4.4641 -.1554 =.3087 z -3 -2 -1 0 1 2 3 -1.8 -.4 Draw and shade area to find a value between 41 and 48. z = x – x s z = 41 – 50 5 z = -1.8 Given a normally distributed population with mean = 50 and standard deviation = 5. Each data item x can be converted to a standard value z by the formula: z = x – x s

10. Given a sample of 600 children where 486 preferred a P B & J sandwich for their lunch.   s = p (1 – p)  √ n 9. Compute p 486 600.81 10. Compute the standard deviation  s = p (1 – p) √ n s =.81(.19) √ 600.016

11.     p – 2 s < p < p + 2s p – 3 s < p < p + 3s 11. Give a 95 % confidence interval for p.   p – 2 s < p < p + 2s.81–2( ) < p <.81 + 2( ).81–2(.016) < p <.81 + 2(.016).778 < p <.842 12.Give a 99 % confidence interval for p.  p – 3 s < p < p + 3s.81 – 3( ) < p <.81 + 3( ).81 – 3(.016) < p <.81 + 3(.016).762 < p <.858

12. Eleven used cars of the same make and model were randomly selected. The gas mileage and # of months since last tune up are noted below. 14. Find the regression equation. y = -.33x + 19.83 16. Compute y. 16.64 15. Find r. -.948 13. Make a scatter plot. TimeMPG 816 1314 319 120 817 1016 15 2014 419 220 2313

13. 17. At the 95 % confidence level, predict the mpg for a car whose last tune-up occurred 17 months ago. n = 11 95% -.948.602 Since r > table, significant so substitute 17 for x in equation y = -.33x + 19.83 y = -.33(17) + 19.83 14.25 mpg 18. At the 99 % confidence level, predict the mpg for a car whose last tune-up occurred 30 months ago. n = 11 99% -.948.735 Since r > table, significant so substitute 30 for x in equation y =-.33x + 19.83 y = -.33(30) + 19.83 10 mpg