the life time of a certain kind of mean life of 400 hours and standard deviation of 45 hours. assuming the distribution of life time to be normal, find : 1. the percentage of batteries with a life time of atleast 470 hours 2. the proportion of batteries with a lifetime between 385 and 415 hours 3. the minimum life of the best 5% of the batteries

Z = (score-mean)/SD

1. Z = (470-400)/45 = ?

2. Use same equation for both values.

3. Find Z score in table and work backwards.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score(s).

Oh, battery life, huh? Well, let's get cracking!

1. To find the percentage of batteries with a life time of at least 470 hours, we'll use some z-scores. First, we need to find the z-score for 470 hours:

z = (x - μ) / σ

Where x is the value we're interested in (470 hours), μ is the mean (400 hours), and σ is the standard deviation (45 hours). Plugging in the values:

z = (470 - 400) / 45

Calculating this, we find that z ≈ 1.56. Now, we need to find the percentage of values that fall to the right of this z-score on the standard normal distribution. From a standard normal distribution table (or using a calculator), we find that this percentage is approximately 6.68%.

So, roughly 6.68% of batteries have a life time of at least 470 hours. Keep chugging along, little batteries!

2. To find the proportion of batteries with a lifetime between 385 and 415 hours, we'll use the same approach. First, we calculate the z-scores for both values:

For 385 hours:

z1 = (385 - 400) / 45

For 415 hours:

z2 = (415 - 400) / 45

Calculating these, we find z1 ≈ -0.33 and z2 ≈ 0.33. Now we need to find the difference between the cumulative probabilities for these two z-scores. Using a standard normal distribution table (or a calculator), we find that the cumulative probability for z1 is approximately 0.3707 and for z2 is approximately 0.6331.

So, the proportion of batteries with a lifetime between 385 and 415 hours is approximately 0.6331 - 0.3707 = 0.2624, or 26.24%. Not too shabby!

3. Now, for the minimum life of the best 5% of batteries! First, we need to find the z-score that corresponds to the 5th percentile. Using a standard normal distribution table, we find that the z-score for the 5th percentile is approximately -1.645.

Now we can use this z-score to find the minimum life by rearranging the formula:

x = z * σ + μ

Plugging in the values:

x = -1.645 * 45 + 400

Calculating this, we find that x ≈ 325.03. So, the minimum life of the best 5% of batteries is approximately 325.03 hours.

Ta-da! Batteries and statistics, what a combo! If you have any more electrifying questions, feel free to ask!

To solve these questions, we need to use the properties of the normal distribution and the formula for Z-score.

1. To find the percentage of batteries with a life time of at least 470 hours, we need to find the area under the normal curve to the right of 470 hours. We can use the z-score formula:

Z = (X - μ) / σ

where:
X = 470 (value we want to find the area beyond)
μ = 400 (mean)
σ = 45 (standard deviation)

Substituting the values into the formula, we find:

Z = (470 - 400) / 45
Z = 1.56

Using a standard normal distribution table or calculator, we can find that the area to the right of 1.56 is approximately 0.0594, which is equivalent to 5.94%. Therefore, the percentage of batteries with a life time of at least 470 hours is approximately 5.94%.

2. To find the proportion of batteries with a lifetime between 385 and 415 hours, we need to find the area under the normal curve between these two values. We can again use the z-score formula:

Z = (X - μ) / σ

where:
X1 = 385 (lower value)
X2 = 415 (upper value)
μ = 400 (mean)
σ = 45 (standard deviation)

Substituting the values into the formula, we find:

Z1 = (385 - 400) / 45
Z1 = -0.33

Z2 = (415 - 400) / 45
Z2 = 0.33

Using a standard normal distribution table or calculator, we can find the area to the right of -0.33 and the area to the right of 0.33. Subtracting the two areas gives us the area between -0.33 and 0.33, which is approximately 0.255. Therefore, the proportion of batteries with a lifetime between 385 and 415 hours is approximately 25.5%.

3. To find the minimum life of the best 5% of the batteries, we need to find the value corresponding to the z-score that encloses 5% of the area to the right. Using a standard normal distribution table or calculator, we can find that the z-score corresponding to the 5% area to the right is approximately 1.645.

Using the z-score formula:

Z = (X - μ) / σ

where:
Z = 1.645 (z-score for the 5% area to the right)
μ = 400 (mean)
σ = 45 (standard deviation)

Substituting the values into the formula and solving for X, we find:

1.645 = (X - 400) / 45
X - 400 = 1.645 * 45
X - 400 = 73.9
X = 400 + 73.9
X ≈ 473.9

Therefore, the minimum life of the best 5% of the batteries is approximately 473.9 hours.

To solve these questions, we will use the properties of a normal distribution and the z-score formula.

1. The percentage of batteries with a life time of at least 470 hours:
To find this percentage, we need to calculate the area under the normal distribution curve to the right of 470 hours. First, we need to convert 470 hours to a z-score using the formula:

z = (x - μ) / σ

where x is the value (470), μ is the mean (400), and σ is the standard deviation (45).

z = (470 - 400) / 45
z = 1.56

Using a z-table or a calculator, we find that the area to the right of 1.56 is approximately 0.0606. To find the percentage, we multiply this value by 100:

Percentage = 0.0606 * 100 = 6.06%

Therefore, approximately 6.06% of batteries would have a life time of at least 470 hours.

2. The proportion of batteries with a lifetime between 385 and 415 hours:
To find this proportion, we need to calculate the area under the normal distribution curve between 385 and 415 hours. First, we need to convert these values to z-scores:

z1 = (385 - 400) / 45 = -0.3333
z2 = (415 - 400) / 45 = 0.3333

Using a z-table or a calculator, we find that the area to the right of -0.3333 is approximately 0.3707, and the area to the right of 0.3333 is also approximately 0.3707. To find the proportion, we subtract these values:

Proportion = 0.3707 - 0.3707 = 0

Therefore, approximately 0% of batteries would have a life time between 385 and 415 hours.

3. The minimum life of the best 5% of the batteries:
To find this value, we need to find the z-score that corresponds to the 5% cutoff on the right tail of the distribution. We can use a z-table or a calculator to find this value.

z = 1.645 (approximately, rounded to 3 decimal places)

Now we can use the z-score formula to find the minimum life:

Minimum Life = μ + (z * σ)
Minimum Life = 400 + (1.645 * 45)
Minimum Life = 477.525

Therefore, the minimum life of the best 5% of the batteries would be approximately 477.525 hours.