The Insurance Institute for Highway Safety conducted tests with crashes of new cars traveling at 6 mi/h. The total cost of the damages for a simple random sample of the tested cars are listed below. Based on these results, is damage of $10,000 unusual? Why or why not? $7448 $4911 $9051 $6374 $4277
Yes, the damage of $10,000 would be unusual. Can you determine why based on the data given?
I'll let you take it from here.
To determine if a damage of $10,000 is unusual, we need to analyze the data provided and consider the measures of central tendency and variability. Let's calculate the mean and standard deviation of the damages.
1. Calculate the mean:
Mean = (7448 + 4911 + 9051 + 6374 + 4277) / 5
Mean = 5201
2. Calculate the standard deviation:
a. Calculate the differences from the mean for each data point:
(7448 - 5201), (4911 - 5201), (9051 - 5201), (6374 - 5201), (4277 - 5201)
2247, -290, 3850, 1173, -924
b. Square each difference:
2247^2, (-290)^2, 3850^2, 1173^2, (-924)^2
5056009, 84100, 14822500, 1374529, 853776
c. Calculate the mean of the squared differences:
Mean of squared differences = (5056009 + 84100 + 14822500 + 1374529 + 853776) / 5
Mean of squared differences = 3761822
d. Take the square root of the mean of squared differences to find the standard deviation:
Standard deviation = √3761822
Standard deviation ≈ 1939
Now, to determine if a damage of $10,000 is unusual, we need to see if it falls within a certain range around the mean. Typically, damages within two standard deviations of the mean are considered usual or normal.
Lower limit = Mean - (2 * Standard deviation)
Lower limit = 5201 - (2 * 1939)
Lower limit ≈ 1323
Upper limit = Mean + (2 * Standard deviation)
Upper limit = 5201 + (2 * 1939)
Upper limit ≈ 9089
Since the damage of $10,000 falls within the range of 1323 to 9089, it is considered usual or not unusual.
To determine whether a damage amount of $10,000 is unusual or not, we need to analyze the data provided and evaluate where it falls within the range of damages.
First, let's calculate the average (mean) and standard deviation of the damages:
1. Calculate the sum of the damages: $7448 + $4911 + $9051 + $6374 + $4277 = $32,061.
2. Divide the sum by the number of data points (5): $32,061 / 5 = $6,412.2 (mean).
Next, calculate the standard deviation:
1. Calculate the difference between each data point and the mean:
- ($7448 - $6,412.2) = $1,035.8
- ($4911 - $6,412.2) = -$1,501.2
- ($9051 - $6,412.2) = $2,638.8
- ($6374 - $6,412.2) = $961.8
- ($4277 - $6,412.2) = -$2,135.2
2. Square each difference:
- ($1,035.8)^2 = $1,072,790.44
- (-$1,501.2)^2 = $2,253,602.44
- ($2,638.8)^2 = $6,963,844.44
- ($961.8)^2 = $925,694.44
- (-$2,135.2)^2 = $4,555,244.04
3. Calculate the average of these squared differences:
($1,072,790.44 + $2,253,602.44 + $6,963,844.44 + $925,694.44 + $4,555,244.04) / 5 = $2,553,835.76
4. Take the square root of this average:
√$2,553,835.76 = $1,597.99 (standard deviation).
Now, to determine if a damage amount of $10,000 is unusual or not, we can use the concept of "Z-scores." The Z-score measures the relative position of a data point within a distribution.
To calculate the Z-score for $10,000, use the formula: Z = (x - mean) / standard deviation.
Z = ($10,000 - $6,412.2) / $1,597.99 = 2.25
A Z-score of 2.25 tells us that the damage amount of $10,000 is 2.25 standard deviations above the mean. In statistical terms, any Z-score greater than 2 (or less than -2) is generally considered unusual or outliers.
Therefore, since the Z-score is 2.25 (greater than 2), we can conclude that a damage amount of $10,000 is considered unusual based on the given dataset.