The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean  u=41  and standard deviation  o=4 . Use the TI-84 Plus calculator to answer the following.

(a) Find the 21st percentile of the tire lifetimes.

(b) Find the 73rd percentile of the tire lifetimes.

(c) Find the first quartile of the tire lifetimes.

(d) The tire company wants to guarantee that its tires will last at least a certain number of miles. What number of miles (in thousands) should the company guarantee so that only 4% of the tires violate the guarantee?

Round the answers to at least two decimal places.

To answer these questions, we will use the **inverse normal function** (`invNorm`) on the TI-84 Plus calculator, which finds the value corresponding to a specific percentile in a normal distribution. The syntax for this function is:

```
invNorm(area to the left, mean, standard deviation)
```

Here, the mean \(\mu = 41\) and the standard deviation \(\sigma = 4\).

### (a) Find the 21st percentile of the tire lifetimes.
The 21st percentile means that 21% of the data is below this value, so we are looking for \( P(X \leq x) = 0.21 \).

**Steps on the TI-84:**
- Press `2nd` then `VARS` to open the **DISTR** menu.
- Choose `3:invNorm(`.
- Enter `0.21, 41, 4` (because we want the 21st percentile, with mean 41 and standard deviation 4).
- Press `ENTER`.

The result is approximately **37.74**.

### (b) Find the 73rd percentile of the tire lifetimes.
The 73rd percentile means that 73% of the data is below this value, so we are looking for \( P(X \leq x) = 0.73 \).

**Steps on the TI-84:**
- Press `2nd` then `VARS` to open the **DISTR** menu.
- Choose `3:invNorm(`.
- Enter `0.73, 41, 4`.
- Press `ENTER`.

The result is approximately **43.10**.

### (c) Find the first quartile of the tire lifetimes.
The first quartile (Q1) corresponds to the 25th percentile, so we are looking for \( P(X \leq x) = 0.25 \).

**Steps on the TI-84:**
- Press `2nd` then `VARS` to open the **DISTR** menu.
- Choose `3:invNorm(`.
- Enter `0.25, 41, 4`.
- Press `ENTER`.

The result is approximately **38.31**.

### (d) The tire company wants to guarantee that its tires will last at least a certain number of miles. What number of miles (in thousands) should the company guarantee so that only 4% of the tires violate the guarantee?
This is asking for the 4th percentile, meaning \( P(X \leq x) = 0.04 \).

**Steps on the TI-84:**
- Press `2nd` then `VARS` to open the **DISTR** menu.
- Choose `3:invNorm(`.
- Enter `0.04, 41, 4`.
- Press `ENTER`.

The result is approximately **34.42**.

### Summary of results:
(a) 21st percentile: **37.74 thousand miles**

(b) 73rd percentile: **43.10 thousand miles**

(c) First quartile (25th percentile): **38.31 thousand miles**

(d) To guarantee that only 4% of the tires violate the guarantee, the company should guarantee at least **34.42 thousand miles**.

---

Let me know if you need further details or explanations. Here are some related questions:

1. How would the results change if the standard deviation was 5 instead of 4?
2. What is the interpretation of a percentile in a normal distribution?
3. How can we calculate probabilities for values greater than a given percentile?
4. What is the significance of the first and third quartiles in a normal distribution?
5. How does changing the mean affect the percentile values?

**Tip**: Always round answers to at least two decimal places when using the TI-84 calculator for normal distribution problems to ensure accuracy.

The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean μ=41 and standard deviation σ=4. Use the TI-84 Plus calculator to answer the following:
(a) Find the 21st percentile of the tire lifetimes.
(b) Find the 73rd percentile of the tire lifetimes.
(c) Find the first quartile of the tire lifetimes.
(d) The tire company wants to guarantee that its tires will last at least a certain number of miles. What number of miles (in thousands) should the company guarantee so that only 4% of the tires violate the guarantee?
Round the answers to at least two decimal places.

Learn how to calculate percentiles for a normally distributed tire lifetime (mean = 41, standard deviation = 4) using the TI-84 Plus calculator. Find the 21st, 73rd percentiles, first quartile, and the mileage guarantee to cover 96% of tire lifetimes. Suitable for Grades 10-12 students.

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