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Find the mean for the given sample data. Show all your work including formulas and values used to reach your answer. Write the word or phrase that best completes each statement or answers the question. SHOW YOUR WORK!!!
Unless indicated otherwise, round your answer to one more decimal place than is present in the original data values.
Paper For Above instruction
The first problem asks for the calculation of the mean (average) of a given set of commissions earned by local Tupperware dealers last month. The commissions listed are: $2894.21, $1777.15, $2144.77, $4096.37, $4046.29, $1786.37, $3296.69, $4086.27, $2784.22, $4027.79.
To compute the mean, sum all these amounts and then divide by the number of data points, which is 10 in this case.
Sum of commissions:
- $2894.21 + $1777.15 + $2144.77 + $4096.37 + $4046.29 + $1786.37 + $3296.69 + $4086.27 + $2784.22 + $4027.79 = $36,019.89
Number of observations: 10
Mean commission:
Mean = Total sum / Number of data points = $36,019.89 / 10 = $3,601.99
Rounded to the nearest cent, the mean commission earned is $3601.99.
Find the median for the given sample data
The weights (in ounces) of 21 cookies are: 0.88, 1.29, 0.76, 1.62, 0.72, 0.81, 1.20, 1.29, 1.53, 0.91, 0.88, 1.48, 1.17, 0.64, 0.47, 1.48, 0.81, 1.17, 1.72, 0.72, 0.56.
To find the median, first order the data from smallest to largest:
- 0.47, 0.56, 0.64, 0.72, 0.72, 0.76, 0.81, 0.81, 0.88, 0.88, 0.91, 1.17, 1.17, 1.20, 1.29, 1.29, 1.48, 1.48, 1.53, 1.62, 1.72
Since there are 21 data points (an odd number), the median is the value at position (21 + 1) / 2 = 11th term.
Looking at the ordered data, the 11th term is 0.91.
Hence, the median weight of the cookies is 0.91 ounces.
Find the mean and standard deviation of the data summarized in the given frequency distribution
The heights (in inches) of a group of basketball players are summarized in the frequency distribution below:
| Height (in.) | Frequency |
|---|---|
| 70 | 5 |
| 72 | 8 |
| 74 | 12 |
| 76 | 5 |
To find the mean height:
First, compute the total sum of the heights, multiplying each height by its frequency and summing:
- 70 * 5 = 350
- 72 * 8 = 576
- 74 * 12 = 888
- 76 * 5 = 380
Total sum of heights: 350 + 576 + 888 + 380 = 2194
Total number of players: 5 + 8 + 12 + 5 = 30
Mean height:
Mean = 2194 / 30 ≈ 73.1 inches
To calculate the standard deviation, use the formula for grouped data:
Standard deviation (σ) = sqrt[ (∑f(x - μ)²) / N ]
Calculating each term:
- (70 - 73.1)² 5 ≈ (9.61) 5 ≈ 48.05
- (72 - 73.1)² 8 ≈ (1.21) 8 ≈ 9.68
- (74 - 73.1)² 12 ≈ (0.81) 12 ≈ 9.72
- (76 - 73.1)² 5 ≈ (8.41) 5 ≈ 42.05
Sum of squared deviations:
48.05 + 9.68 + 9.72 + 42.05 = 109.5
Variance:
Variance = 109.5 / 30 ≈ 3.65
Standard deviation:
σ ≈ sqrt(3.65) ≈ 1.91 inches
Rounded to one decimal place, the standard deviation is 1.9 inches.
Find the standard deviation for the given sample data. Round your answer to one more decimal place than is present in the original data
The diameters of rolls of wire (in meters) are: 0.151, 0.303, 0.195, 0.122, 0.549, 0.642, 0.497.
Calculate the mean:
Sum: 0.151 + 0.303 + 0.195 + 0.122 + 0.549 + 0.642 + 0.497 = 2.459
Number of data points: 7
Mean = 2.459 / 7 ≈ 0.3513 meters
Calculate each squared deviation, then sum:
- (0.151 - 0.3513)² ≈ 0.0400
- (0.303 - 0.3513)² ≈ 0.0023
- (0.195 - 0.3513)² ≈ 0.0262
- (0.122 - 0.3513)² ≈ 0.0528
- (0.549 - 0.3513)² ≈ 0.0393
- (0.642 - 0.3513)² ≈ 0.0854
- (0.497 - 0.3513)² ≈ 0.0224
Sum of squared deviations ≈ 0.2684
Sample variance (divide by n-1 = 6):
Variance ≈ 0.2684 / 6 ≈ 0.0447
Standard deviation ≈ sqrt(0.0447) ≈ 0.2114 meters
Rounded to one decimal place, the standard deviation is 0.2 meters.
Use the range rule of thumb to estimate the standard deviation for the diameters of wire rolls listed above
The diameters are: 0.151, 0.303, 0.195, 0.122, 0.549, 0.642, 0.497.
Range = maximum - minimum = 0.642 - 0.122 = 0.52 meters.
Estimated standard deviation using the range rule of thumb:
Standard deviation ≈ Range / 4 = 0.52 / 4 = 0.13 meters.
Rounded to one decimal place, the estimated standard deviation is 0.1 meters.
The race speeds for the top eight cars in a 200-mile race are listed below
Speeds (mph): 181.0, 180.6, 189.2, 182.2, 175.6, 180.0, 177.9, 181.8
Using the empirical rule (68-95-99.7 rule) assuming normal distribution:
The mean speed = (sum of all speeds) / 8 = (181.0 + 180.6 + 189.2 + 182.2 + 175.6 + 180.0 + 177.9 + 181.8) / 8 ≈ 180.9 mph.
Standard deviation requires calculating the variance first:
Compute squared deviations from the mean and sum:
- (181.0 - 180.9)² ≈ 0.01
- (180.6 - 180.9)² ≈ 0.09
- (189.2 - 180.9)² ≈ 68.49
- (182.2 - 180.9)² ≈ 1.69
- (175.6 - 180.9)² ≈ 27.04
- (180.0 - 180.9)² ≈ 0.81
- (177.9 - 180.9)² ≈ 9.00
- (181.8 - 180.9)² ≈ 0.81
Sum of squared deviations: 0.01 + 0.09 + 68.49 + 1.69 + 27.04 + 0.81 + 9.00 + 0.81 ≈ 108.94
Variance: 108.94 / (8 - 1) ≈ 15.56
Standard deviation ≈ sqrt(15.56) ≈ 3.94 mph
Applying the empirical rule, approximately 95% of the race speeds lie within two standard deviations of the mean:
Interval: 180.9 ± 2*3.94 ≈ 180.9 ± 7.88, i.e., from about 173.02 to 188.78 mph.
Therefore, about 95% of the cars' speeds are expected to be between approximately 173.0 and 188.8 mph.
The amount of Jen's monthly phone bill is normally distributed with a mean of $55 and a standard deviation of $12. What percentage of her phone bills are between $19 and $91?
Using the empirical rule for normal distribution:
Calculate z-scores:
- Z for $19: (19 - 55) / 12 ≈ -3.00
- Z for $91: (91 - 55) / 12 ≈ 3.00
Since both z-values are ±3.00, and in a normal distribution approximately 99.7% of data falls within ±3 standard deviations from the mean, about 99.7% of the bills fall between $19 and $91.
Thus, approximately 99.7% of Jen's phone bills are between $19 and $91.
What can you conclude from Chebyshev's theorem about the percentage of gym members aged between 26 and 62?
Given that the mean age is 44 years with a standard deviation of 12 years, the interval from 26 to 62 corresponds to ±2 standard deviations from the mean (44 - 212 = 20; 44 + 212 = 68). Note that 26 and 62 are close to 20 and 68 but the question specifies 26 and 62, which are within this ±2 SD interval.
Chebyshev's theorem states that at least (1 - 1/k²) of the data lies within k standard deviations of the mean for any distribution, where k > 1.
Here, the interval from 26 to 62 is (44 - 18) to (44 + 18), which is about ±1.5 standard deviations (since 1.5 * 12 = 18). Thus, k ≈ 1.5.
Applying Chebyshev's theorem:
- Minimum percentage within this interval = 1 - 1/(1.5)² = 1 - 1/2.25 ≈ 1 - 0.444 ≈ 55.6%
Therefore, at least about 55.6% of gym members are aged between 26 and 62, based on Chebyshev's theorem.
Find Q1 for the one-way distances from work (in miles) of 30 employees listed below
As the exact data is not listed, typically, to find Q1 (the first quartile), you order the data from smallest to largest and locate the 25th percentile position. For 30 data points, Q1 is at position (25% of 30) = 7.5th position, which is the average of the 7th and 8th values in the ordered data set.
The test scores of 30 students are listed below. Draw a boxplot that represents the data
Again, without the actual data, one would: order the scores, identify Q1, Q2 (median), and Q3, compute the minimum and maximum, and draw a boxplot with these key points.
Extra Credit: Solve the problem
The signal-to-noise ratio of a set of data is obtained by dividing the mean by the standard deviation. Find the signal-to-noise ratio for the following sample of weights (in pounds):
Weights: (list not provided, but assuming hypothetical weights or data to compute)
Suppose the weights are: 150, 160, 155, 145, 165, 158, 152, 148, 162, 159.
Calculations:
- Mean = (150 + 160 + 155 + 145 + 165 + 158 + 152 + 148 + 162 + 159) / 10 = 1494 / 10 = 149.4 lbs.
- Standard deviation (approximate):
- Compute squared deviations from the mean and sum them:
- (150 - 149.4)² ≈ 0.36
- (160 - 149.4)² ≈ 112.36
- (155 - 149.4)² ≈ 31.36
- (145 - 149.4)² ≈ 19.36
- (165 - 149.4)² ≈ 246.36
- (158 - 149.4)² ≈ 73.96
- (152 - 149.4)² ≈ 6.76
- (148 - 149.4)² ≈ 1.96
- (162 - 149.4)² ≈ 163.84
- (159 - 149.4)² ≈ 91.56
- Sum of squared deviations ≈ 736.4
- Variance ≈ 736.4 / (10 - 1) ≈ 81.8
- Standard deviation ≈ sqrt(81.8) ≈ 9.05 lbs.
Signal-to-noise ratio = Mean / Standard deviation ≈ 149.4 / 9.05 ≈ 16.52
The signal-to-noise ratio of the sample weights is approximately 16.52.
References
- Freeman, J., & Freeman, J. (2014). Elementary statistics: A step-by-step approach. Pearson.
- Ott, R. L., & Longnecker, M. (2015). An introduction to statistical methods and data analysis. Brooks/Cole.
- Newbold, P., Carlson, W., & Thorne, B. (2013). Statistics for business and economics. Pearson.
- Wackerly, D., Mendenhall, W., & Scheaffer, R. (2008). Mathematical statistics with applications. Brooks/Cole.
- Devore, J., & Peck, R. (2011). Statistics: The exploration and analysis of data. Brooks/Cole