Instructions for Side by Side Printing
- Print the notecards
- Fold each page in half along the solid vertical line
- Cut out the notecards by cutting along each horizontal dotted line
- Optional: Glue, tape or staple the ends of each notecard together
General Statistics: Ch 6 Quiz
front 1 Assume that thermometer readings are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A thermometer is randomly selected and tested. For the case below, draw a sketch, and find the probability of the reading. (The given values are in Celsius degrees.) Between –1.47 and 1.87 | back 1  The probability of getting a reading between –1.47°C and 1.87°C is 0.8985 . P(-1.47 < z < 1.87) = (0.9693 – 0.0708) |
front 2 Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the probability that a randomly selected thermometer reads greater than 0.04 and draw a sketch of the region. | back 2  The probability is 0.4840.
P(z > 0.04) = 1 – P(z < 0.04) = 1 – 0.5160 = 0.4840 OR Symmetric to z score of -0.04 (Area = 0.4840) |
front 3 Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is less than –0.56 and draw a sketch of the region. | back 3  The probability is 0.2877. P(z < -0.56) = 0.2877 |
front 4  Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. | back 4 The indicated z score is 2.09. |
front 5  Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. | back 5 The indicated z score is –0.54.
Symmetric to Area of 0.7054 (z = 0.54) OR Area = 1 – 0.7054 = 0.2946 |
front 6 A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 49.0 and 59.0 minutes. Find the probability that a given class period runs between 50.5 and 51.5 minutes. | back 6 0. 1 00
49.0 = 10.0 Since the uniform distribution is rectangular, has a length of 10.0, and an area of 1, the height of the uniform distribution is 0.1. P(between 50.5 and 51.5) = (shaded region length) × (shaded region height) = (51.5 – 50.5) × (0.1) = 0.1 |
front 7 Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is less than 3.69 and draw a sketch of the region. | back 7  The probability is 0.9999. |
front 8  Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1. | back 8 The area of the shaded region is 0.8508.
P(z > 0.04) = 1 – P(z < -1.04) = 1 – 0.1492 = 0.8505 OR Symmetric to z score of 1.04 (Area = 0.8508) |
front 9 The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 6 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 4.25 minutes. | back 9 0.298 P(greater than 45.25) = (length of shaded region) × (height of shaded region)
= (6 – 5.25) × (0.167)-round to 2 dec. places = (1.75 × (0.17) = 0.2975 |
front 10 Assume that thermometer readings are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A thermometer is randomly selected and tested. For the case below, draw a sketch, and find the probability of the reading. (The given values are in Celsius degrees.) Between –0.50 and 2.00 | back 10  The probability of getting a reading between 0.50°C and 2.00°C is 0.2857. P(0.50 < z < 2.00) = (0.9772 – 0.6915) |