Distance Breaking The Breaking Distance Of A Simple Random Sample Of Cars: 95% Confidence N = 32; ~ = 137ft And A Is Known To Be 7 Ft.
The correct answer and explanation is :
Let’s first solve the problem correctly:
You are given:
- Sample size, ( n = 32 )
- Sample mean, ( \bar{x} = 137 ) ft
- Population standard deviation, ( \sigma = 7 ) ft
- Confidence level = 95%
Since the population standard deviation ( \sigma ) is known and ( n > 30 ), we use the z-distribution to construct the confidence interval.
The formula for a confidence interval when ( \sigma ) is known:
[
\bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}
]
Where:
- ( \bar{x} ) = sample mean
- ( z_{\alpha/2} ) = z-score corresponding to the desired confidence level
- ( \sigma ) = population standard deviation
- ( n ) = sample size
At 95% confidence, ( z_{\alpha/2} = 1.96 ) (standard value).
Now, calculate the standard error (SE):
[
SE = \frac{\sigma}{\sqrt{n}} = \frac{7}{\sqrt{32}} \approx \frac{7}{5.6569} \approx 1.237
]
Now compute the margin of error (ME):
[
ME = z_{\alpha/2} \times SE = 1.96 \times 1.237 \approx 2.424
]
Finally, construct the confidence interval:
[
137 \pm 2.424
]
Thus:
- Lower limit = ( 137 – 2.424 = 134.576 )
- Upper limit = ( 137 + 2.424 = 139.424 )
Final Answer:
✅ The 95% confidence interval for the mean breaking distance is approximately (134.58 ft, 139.42 ft).
300-word Explanation:
In this problem, we are asked to calculate a 95% confidence interval for the mean breaking distance of cars based on a sample. The sample size provided is 32 cars, and the sample mean distance is 137 feet. Importantly, the standard deviation of the population is given as 7 feet.
Since the standard deviation of the population is known and the sample size is greater than 30, it is appropriate to use the z-distribution instead of the t-distribution. The z-distribution is ideal for large samples with a known standard deviation because it assumes that the sampling distribution of the sample mean is approximately normal.
The formula used to construct a confidence interval is:
[
\bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}
]
where ( \bar{x} ) is the sample mean, ( \sigma ) is the population standard deviation, and ( n ) is the sample size.
First, we calculate the standard error (SE), which measures how much the sample mean is expected to vary from the true population mean. Using ( SE = \sigma / \sqrt{n} ), we find that SE is approximately 1.237. Then, we find the margin of error (ME) by multiplying the SE by the z-score corresponding to a 95% confidence level, which is 1.96.
The resulting margin of error is about 2.424. Adding and subtracting this margin of error from the sample mean gives the confidence interval. Thus, we are 95% confident that the true mean breaking distance for all cars falls between 134.58 feet and 139.42 feet.