Question
Z- score multiple choice Assignment 1. Mr. Green Fatigues is in the 85th percentile in his class with respect to grades. This means, 85% of
Z- score multiple choice Assignment
1. Mr. Green Fatigues is in the 85th percentile in his class with respect to grades. This means,
- 85% of the class is smarter than him.
- He is smarter than 85% of the class.
- He has an 85% average.
- The class average is 85%
2. Suppose a random variable has a z-score of -1.75, a standard deviation of 8 and X = 12. Then the mean of X is,
- -26
- 6
- -6
- 26
- Suppose the masses of Megalodons are normally distributed with a mean of 47,000 kg and a standard deviation of 2,500 kg. Approximately how many, in a group of 175 megalodons, have a mass of more than 51,375 kg?
- 0.9599
- 168
- 167.98
- None of the above.
4. If there are 250 snarks in a colony (a snark is an imaginary animal) and their masses are normally distributed with a mean of 1.8 grams and a standard deviation of 0.5 grams. Approximately how many would have a mass between 1.75 grams and 1.875 grams?
- 0.4009
- All of them
- 100
- 0.5596
5. If we say tat P(Z < a) = 0.9370 this means that z = a is,
- Greater than 6.3 % of other z values
- Less than 93.7% of other z values
- Less than 6.3% of other z values
- Greater than 93.7% of other z values
6. If X~N(15, 25) and x = 17, then x is,
- 1.7 standard deviations above the mean
- 0.4 standard deviations below the mean
- 0.4 standard deviations above the mean
- None of the above
7. Suppose X~N(15, 25), then P(X < 22.5) =
- 0.9332
- 0.9938
- 1.5
- 0.5
8. If P(Z < z) = 0.1075, then z =
- 0.1075
- 0.8925
- 1.24
- -1.24
9. If the mean for X is 50 and the standard deviation for X is 12.5, then z = 1.5 would mean x =
- 55.25
- 51.5
- 68.75
- None of the above
10. The heights of Lawn Trolls are normally distributed with a mean of 35 cm and a standard deviation of 5 cm. If there are 95 Trolls in a lawn, approximately how many are between 32.5 cm and 37.5 cm tall?
- 0.6915
- 36
- 66
- 29
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