The batteries from a certain manufacturer have a mean lifetime of
890
hours, with a standard deviation of
90
hours. Assuming that the lifetimes are normally distributed, complete the following statements.
To complete the statements, we need to calculate the Z-score for a given lifetime value.
The Z-score is calculated by subtracting the mean from the lifetime value and dividing it by the standard deviation.
The formula for the Z-score is:
Z = (x - μ) / σ
where:
Z is the Z-score,
x is the lifetime value,
μ is the mean lifetime,
and σ is the standard deviation.
Statement 1: The Z-score for a battery that lasts 1000 hours is calculated as follows:
Z = (1000 - 890) / 90
Now let's calculate the value:
Z = 110 / 90
Z ≈ 1.22
Therefore, the Z-score for a battery that lasts 1000 hours is approximately 1.22.
Statement 2: The percentage of batteries that last less than 1000 hours can be found using the Z-score in the standard normal distribution table. The table provides the proportion of values that lie below a given Z-score.
Let's consult the standard normal distribution table to find the percentage:
Looking up the Z-score of 1.22 in the standard normal distribution table, we find that approximately 88.87% of the batteries last less than 1000 hours.
Therefore, the percentage of batteries that last less than 1000 hours is approximately 88.87%.
Note: The actual values in the table may differ slightly, so it is always better to use a more accurate source or statistical software to find the exact values.
To complete the statements, we need to provide information about the probabilities associated with the battery lifetimes. Specifically, we need to find the probability of a battery lifetime falling within a certain range.
Statement 1: The probability that a battery, randomly chosen from this manufacturer, will last less than 800 hours.
To find this probability, we need to calculate the area under the normal distribution curve to the left of 800 hours. We can use the Z-score formula to convert the battery lifetime to a standardized value (Z-score).
Z = (X - μ) / σ
Where:
Z is the Z-score
X is the battery lifetime (800 hours in this case)
μ is the mean lifetime (890 hours)
σ is the standard deviation (90 hours)
Plugging in the values, we get:
Z = (800 - 890) / 90 = -1
Now, we can look up the probability associated with a Z-score of -1 in a standard normal distribution table or use a calculator. The probability associated with a Z-score of -1 is approximately 0.1587.
So, the probability that a battery will last less than 800 hours is 0.1587, or 15.87%.
Statement 2: The probability that a battery, randomly chosen from this manufacturer, will last between 750 and 950 hours.
To find this probability, we need to calculate the area under the normal distribution curve between 750 and 950 hours. We can calculate the Z-scores for both these values and then find the area between them.
Z1 = (750 - 890) / 90 = -1.55
Z2 = (950 - 890) / 90 = 0.67
Now, we can look up the probabilities associated with these Z-scores in a standard normal distribution table or use a calculator. The probability associated with a Z-score of -1.55 is approximately 0.0618, and the probability associated with a Z-score of 0.67 is approximately 0.7486.
To find the probability between these two Z-scores, we subtract the probability associated with the smaller Z-score from the probability associated with the larger Z-score:
0.7486 - 0.0618 = 0.6868
So, the probability that a battery will last between 750 and 950 hours is 0.6868, or 68.68%.
By using the Z-score formula and referring to a standard normal distribution table or using a calculator, we can find the probabilities for different ranges of battery lifetimes for a certain manufacturer.
What statements? You cannot copy and paste here.
Just guessing:
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to your Z scores.