A firm produces batteries that have a lifetime which is normally distributed with a mean of 360 minutes and a standard deviation of 30 minutes. The firm needs to keep an eye on the production process to ensure that everything is working properly and that batteries are not being produced that do not meet the advertised standard. This is done by calculating the mean of the sample. To do this they regularly select a sample of 36 batteries in order to test the process.
A State a range within which you would expect the middle 80% of the sample means to lie.
If the process were working correctly, what is the probability that a sample would produce a mean of less than 350 minutes?
B Based on your answer to part (c), what would you conclude about the process if the sample produced a mean life of batteries of 350 minutes?
C What is the probability that two samples in a row would have a mean life time of less than 350 minutes?
I am very lost with these - if anyone can help me with an answer and how they got to that answer?
To answer these questions, we need to use the concept of sampling distributions and the properties of the normal distribution.
A) To find the range within which you would expect the middle 80% of the sample means to lie, we need to find the corresponding z-scores.
Step 1: Find the z-score corresponding to the lower end of the middle 80% range.
To find the lower z-score, we need to find the area to the left of the z-score that leaves 10% in each tail. Since the middle 80% is divided into two tails of 10% each, we subtract 10% from 100% to get 90%. The remaining 90% is divided by 2, resulting in 45% on each tail.
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to a cumulative area of 0.45 is approximately 1.645.
Step 2: Find the z-score corresponding to the upper end of the middle 80% range.
To find the upper z-score, we use the same calculation as above. The cumulative area to the left of the upper z-score will also be 0.45. So, the z-score is also approximately 1.645.
Step 3: Convert the z-scores back to sample means using the formula:
sample mean = population mean + (z-score * standard deviation of sample mean)
The formula for the standard deviation of the sample mean (also known as the standard error) is:
standard deviation of sample mean = standard deviation of population / sqrt(sample size)
Given that the population mean is 360 minutes, the standard deviation of the population is 30 minutes, and the sample size is 36, we can calculate the standard deviation of the sample mean as follows:
standard deviation of sample mean = 30 / sqrt(36) = 5
Now, we can convert the z-scores back to sample means:
Lower sample mean = 360 + (1.645 * 5) = 368.225
Upper sample mean = 360 + (-1.645 * 5) = 351.775
Therefore, we would expect the middle 80% of the sample means to lie within the range of 351.775 to 368.225.
B) To find the probability that a sample will produce a mean of less than 350 minutes, we need to convert 350 minutes to a z-score and then find the corresponding area under the standard normal distribution curve.
The formula to convert a raw score (X) to a z-score is:
z-score = (X - population mean) / standard deviation of population
Using the given values, we calculate the z-score as follows:
z-score = (350 - 360) / 30 = -0.3333
Using a standard normal distribution table or a calculator, you can find the area to the left of this z-score. In this case, the area is approximately 0.3694.
Therefore, the probability that a sample would produce a mean of less than 350 minutes is approximately 0.3694 or 36.94%.
C) To find the probability that two samples in a row would have a mean lifetime of less than 350 minutes, we can use the same approach as in part B. The probability of each sample having a mean lifetime of less than 350 minutes is approximately 0.3694.
To find the probability of both events happening (i.e., two samples in a row having a mean lifetime of less than 350 minutes), we multiply the probabilities together:
0.3694 * 0.3694 = 0.1363
Therefore, the probability that two samples in a row would have a mean lifetime of less than 350 minutes is approximately 0.1363 or 13.63%.
In conclusion:
A) The middle 80% of the sample means is expected to lie within the range of 351.775 to 368.225 minutes.
B) If the sample produced a mean life of 350 minutes, it would have a probability of approximately 0.3694 or 36.94%.
C) The probability that two samples in a row would have a mean lifetime of less than 350 minutes is approximately 0.1363 or 13.63%.
To answer these questions, we can use the Central Limit Theorem (CLT) and the properties of the normal distribution.
A) To find the range within which the middle 80% of the sample means lie, we need to calculate the confidence interval.
Since the population standard deviation is known (30 minutes) and the sample size is sufficiently large (n = 36), we can use the z-distribution.
The z-score corresponding to the middle 80% of the distribution is calculated as follows:
Z = 1.96
Now, let's find the margin of error:
Margin of error = Z * (standard deviation / √sample size)
Substituting the given values:
Margin of error = 1.96 * (30 / √36)
= 1.96 * 5
= 9.8 minutes
Therefore, the range within which you would expect the middle 80% of the sample means to lie is 360 ± 9.8 minutes or (350.2, 369.8) minutes.
B) To find the probability that a sample would produce a mean of less than 350 minutes, we need to standardize the sample mean using the z-score and then look up the corresponding probability on the standard normal distribution table.
The z-score can be calculated as follows:
Z = (sample mean - population mean) / (standard deviation / √sample size)
= (350 - 360) / (30 / √36)
= -10 / 5
= -2
Using the standard normal distribution table, we can find that the probability of a z-score less than -2 is approximately 0.0228 or 2.28%.
Therefore, if the process were working correctly, there is a 2.28% chance that a sample would produce a mean of less than 350 minutes.
C) To find the probability that two samples in a row would have a mean lifetime of less than 350 minutes, we need to calculate the probability of a sample mean less than 350 minutes and then square it for two consecutive samples.
Using the z-score:
Z = (350 - 360) / (30 / √36)
= -10 / 5
= -2
The probability of a sample mean less than 350 minutes can be found using the standard normal distribution table, which is approximately 0.0228 or 2.28%.
The probability of two consecutive samples having a mean lifetime less than 350 minutes is given by squaring this probability:
P = (0.0228)²
= 0.000520224 or 0.052%.
Therefore, the probability that two samples in a row would have a mean lifetime of less than 350 minutes is approximately 0.052%.