a company claims the life of its car batteries is normally distributed with a mean of 3.1 years and a…

Explanation:

Step1: Recall the 68-95-99.7 rule (empirical rule) for normal distribution, which states that for a normal distribution:

• Approximately 68% of the data lies within $\mu \pm \sigma$
• Approximately 95% of the data lies within $\mu \pm 2\sigma$
• Approximately 99.7% of the data lies within $\mu \pm 3\sigma$

where $\mu$ is the mean and $\sigma$ is the standard deviation.

Given $\mu = 3.1$ years and $\sigma = 0.45$ years.

Step2: Calculate the number of standard deviations away from the mean for the values 1.75 and 2.2.

First, for 1.75: Let's find how many standard deviations below the mean 1.75 is. We calculate the difference from the mean: $3.1 - 1.75 = 1.35$ Now, divide by $\sigma$: $\frac{1.35}{0.45} = 3$ So, $1.75 = \mu - 3\sigma$ (since $3.1 - 3\times0.45 = 3.1 - 1.35 = 1.75$)

Next, for 2.2: Find the difference from the mean: $3.1 - 2.2 = 0.9$ Divide by $\sigma$: $\frac{0.9}{0.45} = 2$ So, $2.2 = \mu - 2\sigma$ (since $3.1 - 2\times0.45 = 3.1 - 0.9 = 2.2$)

Step3: Determine the percentage of data between $\mu - 3\sigma$ and $\mu - 2\sigma$.

We know that the total percentage of data within $\mu - 3\sigma$ to $\mu + 3\sigma$ is 99.7%, so the percentage of data below $\mu - 3\sigma$ is $\frac{100 - 99.7}{2} = 0.15%$

The total percentage of data within $\mu - 2\sigma$ to $\mu + 2\sigma$ is 95%, so the percentage of data below $\mu - 2\sigma$ is $\frac{100 - 95}{2} = 2.5%$

To find the percentage of data between $\mu - 3\sigma$ (1.75) and $\mu - 2\sigma$ (2.2), we subtract the percentage below $\mu - 3\sigma$ from the percentage below $\mu - 2\sigma$.

So, the percentage is $2.5% - 0.15% = 2.35%$

Wait, let's check again.

Wait, the empirical rule:

The area between $\mu - 3\sigma$ and $\mu - 2\sigma$ can be found by:

Total area within $\mu - 3\sigma$ to $\mu + 3\sigma$ is 99.7%, so area outside is 0.3%, split equally on both sides: 0.15% below $\mu - 3\sigma$ and 0.15% above $\mu + 3\sigma$.

Total area within $\mu - 2\sigma$ to $\mu + 2\sigma$ is 95%, so area outside is 5%, split equally on both sides: 2.5% below $\mu - 2\sigma$ and 2.5% above $\mu + 2\sigma$.

Therefore, the area between $\mu - 3\sigma$ and $\mu - 2\sigma$ is (percentage below $\mu - 2\sigma$) - (percentage below $\mu - 3\sigma$) = $2.5% - 0.15% = 2.35%$? Wait, no, wait. Wait, actually, the area from $\mu - 3\sigma$ to $\mu - 2\sigma$ is part of the left tail? Wait, no, wait. Wait, the mean is 3.1, so 1.75 is to the left of 2.2, both to the left of the mean.

Wait, let's visualize the normal curve:

Mean at 3.1.

$\mu - 3\sigma = 1.75$, $\mu - 2\sigma = 2.2$, $\mu - \sigma = 3.1 - 0.45 = 2.65$, $\mu = 3.1$, $\mu + \sigma = 3.55$, $\mu + 2\sigma = 4.0$, $\mu + 3\sigma = 4.45$.

So the values 1.75 and 2.2 are both on the left side of the mean.

The area between $\mu - 3\sigma$ and $\mu - 2\sigma$:

We know that the area from $\mu - 3\sigma$ to $\mu + 3\sigma$ is 99.7%, so the area from $\mu - 3\sigma$ to $\mu - 2\sigma$ plus the area from $\mu - 2\sigma$ to $\mu + 2\sigma$ plus the area from $\mu + 2\sigma$ to $\mu + 3\sigma$ is 99.7%.

But we know the area from $\mu - 2\sigma$ to $\mu + 2\sigma$ is 95%, so the area from $\mu - 3\sigma$ to $\mu - 2\sigma$ plus the area from $\mu + 2\sigma$ to $\mu + 3\sigma$ is $99.7% - 95% = 4.7%$. Since the normal distribution is symmetric, the area from $\mu - 3\sigma$ to $\mu - 2\sigma$ is equal to the area from $\mu + 2\sigma$ to $\mu + 3\sigma$. Therefore, each of these areas is $\frac{4.7%}{2} = 2.35%$? Wait, no, wait:

Wait, the area outside $\mu - 2\sigma$ to $\mu + 2\sigma$ is 5%, so the area below $\mu - 2\sigma$ is 2.5%, and the area above $\mu + 2\sigma$ is 2.5%.

The area outside $\mu - 3\sigma$ to $\mu + 3\sigma$ is 0.3%, so the area below $\mu - 3\sigma$ is 0.15%, and the area above $\mu + 3\sigma$ is 0.15%.

Therefore, the area between $\mu - 3\sigma$ and $\mu - 2\sigma$ is (area below $\mu - 2\sigma$) - (area below $\mu - 3\sigma$) = $2.5% - 0.15% = 2.35%$? Wait, but that seems low. Wait, maybe I made a mistake.

Wait, let's recalculate:

Wait, the 68-95-99.7 rule:

• 68% within 1σ of mean: so 32% outside (16% below μ - σ, 16% above μ + σ)
• 95% within 2σ of mean: so 5% outside (2.5% below μ - 2σ, 2.5% above μ + 2σ)
• 99.7% within 3σ of mean: so 0.3% outside (0.15% below μ - 3σ, 0.15% above μ + 3σ)

Therefore, the area between μ - 3σ and μ - 2σ is the area below μ - 2σ minus the area below μ - 3σ, which is 2.5% - 0.15% = 2.35%? Wait, but let's check with the numbers.

Wait, μ = 3.1, σ = 0.45.

μ - 3σ = 3.1 - 3*0.45 = 3.1 - 1.35 = 1.75

μ - 2σ = 3.1 - 2*0.45 = 3.1 - 0.9 = 2.2

So we need the percentage of batteries that last between 1.75 and 2.2 years, which is between μ - 3σ and μ - 2σ.

From the empirical rule, the area between μ - 3σ and μ - 2σ is (area within μ - 3σ to μ + 3σ - area within μ - 2σ to μ + 3σ)/2? Wait, no. Wait, the area within μ - 3σ to μ + 3σ is 99.7%, the area within μ - 2σ to μ + 3σ is area within μ - 2σ to μ + 2σ (95%) plus area within μ + 2σ to μ + 3σ (which is (99.7% - 95%)/2 = 2.35%). Wait, no, that's not right.

Wait, actually, the area from μ - 3σ to μ - 2σ is equal to the area from μ + 2σ to μ + 3σ, because of symmetry.

The total area outside μ - 2σ to μ + 2σ is 5%, so the area below μ - 2σ is 2.5% and above μ + 2σ is 2.5%.

The total area outside μ - 3σ to μ + 3σ is 0.3%, so the area below μ - 3σ is 0.15% and above μ + 3σ is 0.15%.

Therefore, the area between μ - 3σ and μ - 2σ is the area below μ - 2σ minus the area below μ - 3σ, which is 2.5% - 0.15% = 2.35%? Wait, but let's check with a normal distribution table or calculator.

Alternatively, maybe I made a mistake in the number of standard deviations. Let's recalculate the z-scores.

Z-score formula: $z = \frac{x - \mu}{\sigma}$

For x = 1.75:

$z = \frac{1.75 - 3.1}{0.45} = \frac{-1.35}{0.45} = -3$

For x = 2.2:

$z = \frac{2.2 - 3.1}{0.45} = \frac{-0.9}{0.45} = -2$

So we need P(-3 < Z < -2), where Z is the standard normal variable.

From the standard normal table, the cumulative probability for Z = -2 is 0.0228 (2.28%), and for Z = -3 is 0.0013 (0.13%).

Therefore, P(-3 < Z < -2) = P(Z < -2) - P(Z < -3) = 0.0228 - 0.0013 = 0.0215, which is approximately 2.15%, close to our earlier calculation of 2.35% (the difference is due to the empirical rule being approximate, while the standard normal table gives more precise values). However, using the empirical rule, we can say that the percentage is approximately 2.35%, but let's check the empirical rule again.

Wait, the empirical rule states that:

• 68% within μ ± σ (z = -1 to z = 1)
• 95% within μ ± 2σ (z = -2 to z = 2)
• 99.7% within μ ± 3σ (z = -3 to z = 3)

So the area between z = -3 and z = -2 is (99.7% - 95%)/2 = (4.7%)/2 = 2.35%, which matches our first calculation. The slight difference with the standard normal table is because the empirical rule is an approximation.

Therefore, the percentage of the company's car batteries that will last between 1.75 and 2.2 years is approximately 2.35%, but let's confirm.

Wait, let's do the calculation again:

Area between z = -3 and z = -2:

Total area from z = -3 to z = 3: 99.7%

Total area from z = -2 to z = 3: area from z = -2 to z = 2 (95%) + area from z = 2 to z = 3 (which is (99.7% - 95%)/2 = 2.35%)

Therefore, area from z = -3 to z = -2 is equal to area from z = 2 to z = 3, which is 2.35%.

Yes, so the answer is 2.35%? Wait, but let's check with the numbers.

Wait, μ = 3.1, σ = 0.45.

1.75 is 3.1 - 3*0.45 = 1.75

2.2 is 3.1 - 2*0.45 = 2.2

So between 1.75 (μ - 3σ) and 2.2 (μ - 2σ).

From the empirical rule, the percentage between μ - 3σ and μ - 2σ is (99.7% - 95%)/2 = 2.35%.

So the answer is 2.35%? Wait, but maybe the question expects 2.5% - 0.15% = 2.35%, so we can write 2.35 or maybe 2.5 - 0.15 = 2.35.

Wait, but let's check with the standard normal distribution:

The cumulative probability for z = -2 is 0.02275 (from standard normal table), and for z = -3 is 0.00135.

So the probability between z = -3 and z = -2 is 0.02275 - 0.00135 = 0.0214, which is approximately 2.14%, close to 2.35%. The difference is because the empirical rule is an approximation. However, in many cases, the empirical rule is used, so the answer is 2.35% or 2.5% - 0.15% = 2.35%.

Wait, maybe I made a mistake in the z-score calculation. Let's recalculate the z-scores:

For x = 1.75:

z = (1.75 - 3.1)/0.45 = (-1.35)/0.45 = -3. Correct.

For x = 2.2:

z = (2.2 - 3.1)/0.45 = (-0.9)/0.45 = -2. Correct.

So we need P(-3 < Z < -2).

From the empirical rule, the area between z = -3 and z = -2 is (99.7% - 95%)/2 = 2.35%.

Therefore, the percentage is 2.35%.

Answer:

2.35