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

1. Determine the area under the standard normal curve that lies to the right of th type of problem – look up on table z & add to =.8643

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

3. Determine the area under the standard normal curve that lies between -.38 and rd type of problem – look up on table z & add =.5329

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

5. Determine the area under the standard normal curve that lies between.81 and th type of problem – look up on table.2910 z & subtract.4131 –.2910 =.1221

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 –

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 – –.4452 =.0548.

z = x – x s z = 48 – 50 5 z = =.3087 z 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

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 Compute the standard deviation  s = p (1 – p) √ n s =.81(.19) √

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

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 Compute y Find r Make a scatter plot. TimeMPG

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