{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solutionPQ1

# solutionPQ1 - 1 B The tail points to the lower numbers 2 C...

This preview shows pages 1–4. Sign up to view the full content.

1. B. The tail points to the lower numbers 2. C. About half the scores are lower than 155 and half are higher. So, the midpoint of the distribution is close to 155. 3. A. Most of those scores are higher than 120, except 94. 4. B. We should add together all the students who took longer than 90 min to finish the test. About 6 took between 90 ~105 and about 4 took between 105~120. The total should be around 10. 5. B. The balancing point of the distribution is between 45~60. The center should be between 45~60 min. 6. A. The data is slightly skewed right. So the mean will be higher than the median. 7. C. Pos.= (n+1)/2=20/2=10. 75 is at position 10. The median is 75. 8. C. IQR= Q3-Q1= 83-62=21. 9. A. Use your calculator. Instructions as : http://www.geocities.com/calculatorhelp/index.html 10. B Since it follows the standard normal distribution, the standard deviation(s) equals to one. S 2 = variance = 1. 11. A. Z= (42-50)/10=-0.8. Enter table A under Z=-0.8. The table entry is 0.2119. This is the area to the left of –0.8. Because the total area under the curve is 1, the area lying to the right of –0.8 is 1-0.2119=0.7881. 12. A. Standardize first, and we can get –1.7 < x < -0.2. Then, look at the table, the total area should be .4207 – .0446= .3761 13. D. We want to find the score x with area 0.03 to its right under the normal curve. That’s the same as finding the score x with area 0.97 to its left. Use the table A for the entry close to 0.97. It is 1.88. Then, unstandardize it, x = 50 + 1.88*10=68.8.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
14. C. The entry corresponding to z= -1.34 is 0.0901 (this is the probability less), so, this shows most of students’ scores ( 90%) are higher than you. 15. C Z= (48-60)/12=-1 16. B. Look the table A, the entry corresponding to the z-score is 0.1587. 17. A. Unstandardize the z-score. X= 60+12*5.2=122.4. 122.4 is much larger than the mean( =60). 18. B. Range= max- min. And here, the first quartile is bigger than the min, so the IQR = Q3-Q1 is smaller than the range. 19. A. For example, N~(2, 20) is also a normal distribution. The mean can be positive, negative or zero. 20. D. Because the variance involves squaring the deviations, it does not have the same unit of measurement as the original observations. However, the standard deviation s measures spread about the mean in the original scale. 21. A. Z = (9200-7500)/ 1500=1.13. Enter table A under Z=1.13. The table entry is 0.8708. This is the area to the left of 1.13. Because the total area under the curve is 1, the area lying to the right of 1.13 is 1-0.8708=0.1292 22. C. Standardize first, and we can get –1.4 < x < .47. Then, look at the table, the total area should be .6808 – .0808= .6 23. D. So, for x = 4500, you find that the z-score = (4500 - 7500) / 1500 = -2 Then for x= 9000, you find that the z-score = (9000 - 7500) / 1500 = 1 The probability less than -2 equals 0.0228. The probability less than 1 equals 0.8413 . So, the probability is 0.8413-0.0228 = 0.8185 or about 81.5% is between 4500 and 9000 24. B. Since the median is smaller than the mean, we can conclude it’s a skewed right graph. 25. B. Since it is a skewed distribution, the median is better than mean.
26. B. For the skewed right graphs, the mean is farther out in the long tail than is the median. So, the median is less than the mean.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 12

solutionPQ1 - 1 B The tail points to the lower numbers 2 C...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online