Question

1. state the reason why the following frequency distribution is incorrectly constructed.

class frequency
48-53 2
54-59 0
60-65 6
66-72 4
73-78 3
a) a class has been omitted
b) class width is not uniform
c) there is no percent column
d) class limits overlap

1. the following distribution is not a probability distribution because

x -2 -1 0 1 2
p(x) 0.10 0.24 0.41 0.15 0.28
a) the probability values are not increasing.
b) the values of the variable are negative.
c) the probability values do not add to 1.
d) the probability values are not discrete.

Explanation:

Question 5

Step1: Analyze class limits

Check the class limits: 48 - 53, 54 - 59, 60 - 65, 66 - 72, 73 - 78. The class width between 48 - 53 is \(53 - 48 + 1 = 6\) (if inclusive), but wait, actually for continuous (or grouped) data, the class width should be consistent. Wait, no, let's check the gap between classes. Wait, 53 to 54: the previous class ends at 53, next starts at 54, so no overlap. Wait, no, wait the class widths: 53 - 48 = 5? Wait 48 - 53: 53 - 48 = 5? Wait 54 - 59: 59 - 54 = 5? 60 - 65: 65 - 60 = 5? 66 - 72: 72 - 66 = 6? Oh! 66 - 72: 72 - 66 = 6, while others are 5 (48 - 53: 53 - 48 = 5, 54 - 59: 5, 60 - 65: 5). So class width is not uniform. Wait, but let's check again. Wait 48 - 53: 53 - 48 + 1 = 6 (if inclusive), but 54 - 59: 59 - 54 + 1 = 6? Wait no, maybe it's exclusive. Wait, maybe I miscalculated. Wait 48 - 53: upper limit 53, next lower limit 54: so the width is 54 - 48 = 6? Wait 54 - 59: 59 - 54 = 5? No, 59 - 54 = 5? Wait 54 to 59 is 5 units? 59 - 54 = 5? Wait 54,55,56,57,58,59: that's 6 numbers. Wait, maybe the class width is calculated as upper limit - lower limit + 1 for inclusive. Wait 48 - 53: 53 - 48 + 1 = 6. 54 - 59: 59 - 54 + 1 = 6. 60 - 65: 65 - 60 + 1 = 6. 66 - 72: 72 - 66 + 1 = 7? Wait no, 72 - 66 = 6, +1 is 7? Wait, no, I think I messed up. Wait the problem is about frequency distribution. The key is class width uniformity. Wait the classes are 48 - 53, 54 - 59, 60 - 65, 66 - 72, 73 - 78. Let's check the difference between lower limits: 54 - 48 = 6, 60 - 54 = 6, 66 - 60 = 6, 73 - 66 = 7. Oh! 73 - 66 = 7, while others are 6. Wait no, 48 to 54: 6, 54 to 60: 6, 60 to 66: 6, 66 to 73: 7. Wait, no, the classes are 48 - 53 (width 5), 54 - 59 (width 5), 60 - 65 (width 5), 66 - 72 (width 6), 73 - 78 (width 5). Wait, 66 - 72: 72 - 66 = 6, others: 53 - 48 = 5, 59 - 54 = 5, 65 - 60 = 5, 78 - 73 = 5. So the class width for 66 - 72 is 6, others are 5. So class width is not uniform. So option B.

Step2: Evaluate other options

• Option A: No class omitted, as the classes are consecutive (48 - 53, 54 - 59, etc., with no gap between 53 - 54, 59 - 60, 65 - 66, 72 - 73? Wait no, 65 to 66: 65 and 66, so the previous class ends at 65, next starts at 66: no gap. Wait 72 to 73: previous ends at 72, next starts at 73: no gap. So no class omitted.
• Option C: A percent column is not required for a frequency distribution, so this is not a reason for incorrect construction.
• Option D: Class limits: 48 - 53, 54 - 59: no overlap (53 < 54), 59 < 60, 65 < 66, 72 < 73: no overlap. So D is incorrect.

Step1: Recall probability distribution rules

A probability distribution must have two properties: 1) Each \(P(X)\) is between 0 and 1, 2) The sum of all \(P(X)\) is 1.

Step2: Calculate sum of probabilities

Sum the \(P(X)\) values: \(0.10 + 0.24 + 0.41 + 0.15 + 0.28\). Let's calculate: \(0.10 + 0.24 = 0.34\); \(0.34 + 0.41 = 0.75\); \(0.75 + 0.15 = 0.90\); \(0.90 + 0.28 = 1.18\). Wait, no, wait the values: \(0.10\) (for \(X=-2\)), \(0.24\) (\(X=-1\)), \(0.41\) (\(X=0\)), \(0.15\) (\(X=1\)), \(0.28\) (\(X=2\)). Sum: \(0.10 + 0.24 = 0.34\); \(0.34 + 0.41 = 0.75\); \(0.75 + 0.15 = 0.90\); \(0.90 + 0.28 = 1.18\). Wait, that's more than 1. Wait, no, maybe I added wrong. Wait \(0.10 + 0.24 = 0.34\); \(0.34 + 0.41 = 0.75\); \(0.75 + 0.15 = 0.9\); \(0.9 + 0.28 = 1.18\). Wait, that's 1.18, which is not 1. Wait, but wait the problem says "the following distribution is not a probability distribution because". Wait, maybe I misread the values. Wait the table: \(X\): -2, -1, 0, 1, 2; \(P(X)\): 0.10, 0.24, 0.41, 0.15, 0.28. Sum: \(0.10 + 0.24 = 0.34\); \(0.34 + 0.41 = 0.75\); \(0.75 + 0.15 = 0.9\); \(0.9 + 0.28 = 1.18\). Wait, that's over 1. Wait, no, maybe a typo? Wait no, the options: option C says "the probability values do not add to 1". Let's check again. Wait \(0.10 + 0.24 = 0.34\); \(0.34 + 0.41 = 0.75\); \(0.75 + 0.15 = 0.9\); \(0.9 + 0.28 = 1.18\). Yes, sum is 1.18 ≠ 1. So option C.

Step3: Evaluate other options

• Option A: Probability values can be in any order, increasing or not, so this is not a requirement.
• Option B: Negative \(X\) values are allowed in a probability distribution (e.g., a random variable can take negative values), so this is not a reason.
• Option D: Probability values are discrete (each \(P(X)\) is a single value, and \(X\) is discrete), so D is incorrect.

Answer:

B) class width is not uniform

Question 6