The life time of a certain kind of batteries has a mean life of 400 hours and standard deviation as 45 hours. Assuming the distribution of life to be normal, find i) the percentage of batteries with a life time of atleast 470 hours. Ii) Out of 10000 batteries how many batteries do you expect to have life between 385 and 415 hours.
This page will replace any tables or charts in the back of texts or handouts you might have been given
http://davidmlane.com/hyperstat/z_table.html
enter:
mean -- 400
SD -- 45
click on "above" and enter 470
prob is .0599 ---> 5.99%
ii) click on "between" and enter the values
multiply the prob by 10000
The beauty of this page is that you don't have to find z-scores first.
If your course requires you to work with z-scores, then find those scores for all your numbers, and set the mean to 0 and the SD to 1, then enter the data as z-scores.
You will get the same results.
Oh, batteries! They're like little energy ninjas. Let's crunch some numbers!
i) To find the percentage of batteries with a lifetime of at least 470 hours, we'll have to calculate the z-score first. The formula for z-score is: (X - mean) / standard deviation. In this case, X = 470, mean = 400, and standard deviation = 45.
Z-score = (470 - 400) / 45 = 1.56
Now, we need to find the probability associated with this z-score. Using a z-table, we can find that the probability for a z-score of 1.56 is approximately 0.9406.
So, the percentage of batteries with a lifetime of at least 470 hours is 94.06%.
ii) To determine how many batteries we can expect to have a life between 385 and 415 hours out of 10,000 batteries, we'll first find the z-scores for each value.
For 385 hours:
Z-score = (385 - 400) / 45 = -0.33
For 415 hours:
Z-score = (415 - 400) / 45 = 0.33
Using the z-table, we find that the probability for a z-score of -0.33 is approximately 0.3707, and the probability for a z-score of 0.33 is also approximately 0.3707.
To find the percentage in this range, we subtract the lower probability from the upper probability:
0.3707 - 0.3707 = 0
Uh-oh! It seems the probability is zero, which means we don't expect to have any batteries in the range of 385 to 415 hours out of 10,000. Maybe those batteries decided to take a break together.
To find the answers, we can use the properties of the normal distribution.
i) The percentage of batteries with a lifetime of at least 470 hours can be found by calculating the area under the normal curve to the right of 470 hours.
Step 1: Calculate the z-score, which measures how many standard deviations an observation is from the mean. The z-score formula is:
z = (x - mean) / standard deviation
In this case, x = 470, mean = 400, and standard deviation = 45.
So, z = (470 - 400) / 45 = 1.56 (approximately)
Step 2: Use a standard normal distribution table or a calculator with a normal distribution function to find the area to the right of the z-score. The area to the right of 1.56 can be denoted as P(Z > 1.56).
Using a standard normal distribution table or a calculator, we find that P(Z > 1.56) = 0.0594 (approximately).
Step 3: Convert the probability into percentage:
Percentage = 0.0594 × 100 = 5.94%
Therefore, approximately 5.94% of batteries have a lifetime of at least 470 hours.
ii) For this part, we need to find the area under the normal curve between 385 and 415 hours.
Step 1: Calculate the z-scores for both values of x.
For x = 385:
z1 = (385 - 400) / 45 ≈ -0.333
For x = 415:
z2 = (415 - 400) / 45 ≈ 0.333
Step 2: Use symmetry to find the area to the right of -0.333 and the area to the right of 0.333.
P(Z < -0.333) = P(Z > 0.333) ≈ 0.368 (approximately)
Step 3: Calculate the area between -0.333 and 0.333 by subtracting the area to the right of 0.333 from 1.
Area between -0.333 and 0.333 = 1 - (P(Z > 0.333) + P(Z < -0.333)) ≈ 1 - (0.368 + 0.368) ≈ 0.264
Step 4: Convert the probability into a count by multiplying it by the total number of batteries.
Count = 0.264 × 10,000 = 2,640 (approximately)
Therefore, we can expect around 2,640 batteries to have a lifetime between 385 and 415 hours out of 10,000 batteries.
To find the answers to these questions, we need to use the properties of the normal distribution. The normal distribution is defined by its mean (μ) and standard deviation (σ).
In this case, we have the mean life of batteries as 400 hours and the standard deviation as 45 hours.
i) To find the percentage of batteries with a lifetime of at least 470 hours, we need to find the area under the normal distribution curve to the right of 470 hours. We can use the Z-score to calculate this.
The Z-score is calculated using the formula:
Z = (X - μ) / σ
where X is the value we want to find the percentage for (470 hours), μ is the mean (400 hours), and σ is the standard deviation (45 hours).
Z = (470 - 400) / 45
Z = 70 / 45
Z ≈ 1.56
To find the percentage, we can look this Z-score up in a standard normal distribution table or use a calculator. The percentage of batteries with a lifetime of at least 470 hours is the area to the right of the Z-score.
Using a standard normal distribution table or calculator, we find that the area to the right of Z = 1.56 is approximately 0.0594 or 5.94%. Therefore, about 5.94% of batteries have a lifetime of at least 470 hours.
ii) To find the number of batteries expected to have a lifetime between 385 and 415 hours out of 10000 batteries, we need to find the area under the normal distribution curve between these two values.
First, we need to calculate the Z-scores for each value.
Z1 = (385 - 400) / 45
Z1 ≈ -0.333
Z2 = (415 - 400) / 45
Z2 ≈ 0.333
We can then find the area between these Z-scores using a standard normal distribution table or calculator.
The area between Z = -0.333 and Z = 0.333 is approximately 0.2812 or 28.12%.
Therefore, out of 10000 batteries, we can expect approximately 28.12% of them to have a lifetime between 385 and 415 hours. To find the actual number, we can multiply the percentage by the total number of batteries:
Number of batteries = Percentage / 100 * Total
Number of batteries = 28.12 / 100 * 10000
Number of batteries ≈ 2812
So, out of 10000 batteries, we can expect approximately 2812 of them to have a lifetime between 385 and 415 hours.