The distribution of the time taken to run 1000 metres by users of a treadmill in a sports centre has mean 7.2 minutes and standard deviation
2.045 minutes.
1 Choose the option that is closest to the probability that the mean time taken to run 1000 metres for a sample of 30 users will be more than 8 minutes.
Options for Question 1
A 0.0160 B 0.2000 C 0.2486 D 0.8003 E 0.8030 F 0.8106
2 Choose the option that gives an approximate range of values,in minutes, symmetric about the mean, within which the meantime taken to run 1000 metres for approximately 95% of samples of 30 users will lie.
Options for Question 2
A (6.584, 7.816) B (6.469, 7.931) C (3.192, 11.208) D (6.640, 7.760) E (7.066, 7.334) F (4.088, 11.312)
To solve these problems, we need to use the properties of the normal distribution. Given the mean and standard deviation of the distribution of the time taken to run 1000 meters by users of a treadmill, we can use these parameters to determine the probabilities and ranges of values.
1. To find the probability that the mean time taken to run 1000 meters for a sample of 30 users will be more than 8 minutes, we can use the Central Limit Theorem. According to the Central Limit Theorem, for a large sample size, the distribution of the sample means will be approximately normally distributed, regardless of the shape of the original distribution.
The mean of the sample means will be equal to the population mean, which is 7.2 minutes. The standard deviation of the sample means, also known as the standard error of the mean, can be calculated by dividing the standard deviation of the population by the square root of the sample size:
Standard Error of the Mean = Standard Deviation of Population / √(Sample Size)
In this case, the standard deviation of the population is 2.045 minutes, and the sample size is 30. So the standard error of the mean is:
Standard Error of the Mean = 2.045 / √(30) ≈ 0.3747
Now, we need to find the z-score corresponding to the value of 8 minutes using the formula:
z = (x - μ) / σ
where x is the value we are interested in, μ is the mean, and σ is the standard deviation. Plugging in the values:
z = (8 - 7.2) / 0.3747 ≈ 2.13
Using a standard normal distribution table (or a calculator), we can find the cumulative probability associated with a z-score of 2.13. The closest option to the calculated probability is option E: 0.8030.
2. To find an approximate range of values within which the mean time taken to run 1000 meters for approximately 95% of samples of 30 users will lie, we can use the z-score associated with a 95% confidence level, which is 1.96. The range of values can be calculated by multiplying the standard error of the mean by the z-score and then adding/subtracting the result from the mean.
Range of Values = Mean ± (Standard Error of the Mean * Z-Score)
Using the standard error of the mean calculated earlier (0.3747) and the z-score of 1.96, we can calculate the range as:
Range of Values = 7.2 ± (0.3747 * 1.96) ≈ (6.469, 7.931)
So the approximate range of values, in minutes, within which the mean time taken to run 1000 meters for approximately 95% of samples of 30 users will lie is option B: (6.469, 7.931).