Math / Data & Statistics

Question3. Dumbop Tires manufactures a tire with a lletime that approximately follows a nowmal ilistribution with a mean of $70,000 \mathrm{miles}$ and a standari deviation of 4400 mifles, a. What proporion of the tires will last for at least 75,000 miles? b. Sumpose that Dumlop warrants the tires for 60,000 miles. What proportion of tires will last 60,000 miles or less? c. What is the probability that a randomly selected tire lasts between 65,000 and 80,000 miles? d. Suppose that Dunlop wants to warrant no more than $2 \%$ of its tires. What mileage shomld the company advertise as its warranty milcage?

Studdy Solution

STEP 1

What is this asking? We need to figure out how many tires will last for certain mileages, given how long they usually last and how much that varies.
We also need to figure out a good warranty mileage if we only want to replace a small percentage of tires. Watch out! Remember that the normal distribution is all about how likely different values are, and we need to use the mean and standard deviation correctly to get the right answers!

STEP 2

1. At least 75,000 miles
2. 60,000 miles or less
3. Between 65,000 and 80,000 miles
4. Warranty for 2% of tires

STEP 3

Let's define our mean $\mu = 70,000$ and standard deviation $\sigma = 4,400$ .
We want to find the probability that a tire lasts at least 75,000 miles.

STEP 4

To do this, we need to calculate the z-score: $z = \frac{x - \mu}{\sigma} = \frac{75,000 - 70,000}{4,400} \approx 1.14$ This tells us how many standard deviations 75,000 is away from the mean.

STEP 5

Now, we look up this z-score in a z-table (or use a calculator) to find the probability of a tire lasting less than 75,000 miles.
Let's say we find this value to be approximately 0.8729.

STEP 6

Since we want the probability of a tire lasting at least 75,000 miles, we subtract the value we just found from 1: $1 - 0.8729 = 0.1271$ So, about 12.71% of tires will last at least 75,000 miles!

STEP 7

We use the same mean and standard deviation as before.
This time, we want the probability of a tire lasting 60,000 miles or less.

STEP 8

Calculate the z-score: $z = \frac{60,000 - 70,000}{4,400} \approx -2.27$

STEP 9

Looking up this z-score in the z-table, we find the probability is approximately 0.0116.
So, only about 1.16% of tires will last 60,000 miles or less.

STEP 10

We need to find the probability of a tire lasting between 65,000 and 80,000 miles.

STEP 11

Calculate the z-scores for both mileages: $z_1 = \frac{65,000 - 70,000}{4,400} \approx -1.14$ $z_2 = \frac{80,000 - 70,000}{4,400} \approx 2.27$

STEP 12

Look up both z-scores in the z-table.
Let's say we find the probabilities to be approximately 0.1271 for $z_1$ and 0.9884 for $z_2$ .

STEP 13

Subtract the smaller probability from the larger one: $0.9884 - 0.1271 = 0.8613$ About 86.13% of tires will last between 65,000 and 80,000 miles!

STEP 14

We want to find the mileage where only 2% of tires will fail.
This means we're looking for a z-score that corresponds to a probability of 0.02.

STEP 15

Looking in the z-table (or using a calculator), we find the z-score closest to 0.02 is approximately -2.05.

STEP 16

Now, we use the z-score formula to find the mileage: $x = \mu + z \cdot \sigma = 70,000 + (-2.05) \cdot 4,400 \approx 60,980$ So, Dunlop should advertise a warranty mileage of around 60,980 miles!

STEP 17

a. Approximately 12.71% of tires will last at least 75,000 miles. b. Approximately 1.16% of tires will last 60,000 miles or less. c. Approximately 86.13% of tires will last between 65,000 and 80,000 miles. d. Dunlop should offer a warranty of approximately 60,980 miles.

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Studdy Solution

STEP 1

STEP 2

1. At least 75,000 miles
2. 60,000 miles or less
3. Between 65,000 and 80,000 miles
4. Warranty for 2% of tires

STEP 3

Let's define our mean $\mu = 70,000$ and standard deviation $\sigma = 4,400$ .
We want to find the probability that a tire lasts at least 75,000 miles.

STEP 4

To do this, we need to calculate the z-score: $z = \frac{x - \mu}{\sigma} = \frac{75,000 - 70,000}{4,400} \approx 1.14$ This tells us how many standard deviations 75,000 is away from the mean.

STEP 5

Now, we look up this z-score in a z-table (or use a calculator) to find the probability of a tire lasting less than 75,000 miles.
Let's say we find this value to be approximately 0.8729.

STEP 6

Since we want the probability of a tire lasting at least 75,000 miles, we subtract the value we just found from 1: $1 - 0.8729 = 0.1271$ So, about 12.71% of tires will last at least 75,000 miles!

STEP 7

We use the same mean and standard deviation as before.
This time, we want the probability of a tire lasting 60,000 miles or less.

STEP 8

Calculate the z-score: $z = \frac{60,000 - 70,000}{4,400} \approx -2.27$

STEP 9

Looking up this z-score in the z-table, we find the probability is approximately 0.0116.
So, only about 1.16% of tires will last 60,000 miles or less.

STEP 10

We need to find the probability of a tire lasting between 65,000 and 80,000 miles.

STEP 11

Calculate the z-scores for both mileages: $z_1 = \frac{65,000 - 70,000}{4,400} \approx -1.14$ $z_2 = \frac{80,000 - 70,000}{4,400} \approx 2.27$

STEP 12

Look up both z-scores in the z-table.
Let's say we find the probabilities to be approximately 0.1271 for $z_1$ and 0.9884 for $z_2$ .

STEP 13

Subtract the smaller probability from the larger one: $0.9884 - 0.1271 = 0.8613$ About 86.13% of tires will last between 65,000 and 80,000 miles!

STEP 14

We want to find the mileage where only 2% of tires will fail.
This means we're looking for a z-score that corresponds to a probability of 0.02.

STEP 15

Looking in the z-table (or using a calculator), we find the z-score closest to 0.02 is approximately -2.05.

STEP 16

Now, we use the z-score formula to find the mileage: $x = \mu + z \cdot \sigma = 70,000 + (-2.05) \cdot 4,400 \approx 60,980$ So, Dunlop should advertise a warranty mileage of around 60,980 miles!

STEP 17

a. Approximately 12.71% of tires will last at least 75,000 miles. b. Approximately 1.16% of tires will last 60,000 miles or less. c. Approximately 86.13% of tires will last between 65,000 and 80,000 miles. d. Dunlop should offer a warranty of approximately 60,980 miles.

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Studdy Solution

STEP 1

STEP 2

1. At least 75,000 miles2. 60,000 miles or less3. Between 65,000 and 80,000 miles4. Warranty for 2% of tires

STEP 3

Let's **define** our **mean** μ=70,000\mu = 70,000μ=70,000 and **standard deviation** σ=4,400\sigma = 4,400σ=4,400.We want to find the probability that a tire lasts at least **75,000 miles**.

STEP 4

To do this, we need to **calculate** the **z-score**: z=x−μσ=75,000−70,0004,400≈1.14 z = \frac{x - \mu}{\sigma} = \frac{75,000 - 70,000}{4,400} \approx 1.14 z=σx−μ​=4,40075,000−70,000​≈1.14 This tells us how many standard deviations **75,000** is away from the **mean**.

STEP 5

Now, we look up this **z-score** in a **z-table** (or use a calculator) to find the probability of a tire lasting *less* than **75,000 miles**.Let's say we find this value to be approximately **0.8729**.

STEP 6

Since we want the probability of a tire lasting *at least* **75,000 miles**, we subtract the value we just found from **1**: 1−0.8729=0.1271 1 - 0.8729 = 0.1271 1−0.8729=0.1271 So, about **12.71%** of tires will last at least **75,000 miles**!

STEP 7

We use the same **mean** and **standard deviation** as before.This time, we want the probability of a tire lasting **60,000 miles** or less.

STEP 8

**Calculate** the **z-score**: z=60,000−70,0004,400≈−2.27 z = \frac{60,000 - 70,000}{4,400} \approx -2.27 z=4,40060,000−70,000​≈−2.27

STEP 9

Looking up this **z-score** in the **z-table**, we find the probability is approximately **0.0116**.So, only about **1.16%** of tires will last **60,000 miles** or less.

STEP 10

We need to find the probability of a tire lasting between **65,000** and **80,000 miles**.

STEP 11

**Calculate** the **z-scores** for both mileages: z1=65,000−70,0004,400≈−1.14 z_1 = \frac{65,000 - 70,000}{4,400} \approx -1.14 z1​=4,40065,000−70,000​≈−1.14 z2=80,000−70,0004,400≈2.27 z_2 = \frac{80,000 - 70,000}{4,400} \approx 2.27 z2​=4,40080,000−70,000​≈2.27

STEP 12

Look up both **z-scores** in the **z-table**.Let's say we find the probabilities to be approximately **0.1271** for z1z_1z1​ and **0.9884** for z2z_2z2​.

STEP 13

Subtract the smaller probability from the larger one: 0.9884−0.1271=0.8613 0.9884 - 0.1271 = 0.8613 0.9884−0.1271=0.8613 About **86.13%** of tires will last between **65,000** and **80,000 miles**!

STEP 14

We want to find the mileage where only **2%** of tires will fail.This means we're looking for a **z-score** that corresponds to a probability of **0.02**.

STEP 15

Looking in the **z-table** (or using a calculator), we find the **z-score** closest to **0.02** is approximately **-2.05**.

STEP 16

Now, we use the **z-score formula** to find the mileage: x=μ+z⋅σ=70,000+(−2.05)⋅4,400≈60,980 x = \mu + z \cdot \sigma = 70,000 + (-2.05) \cdot 4,400 \approx 60,980 x=μ+z⋅σ=70,000+(−2.05)⋅4,400≈60,980 So, Dunlop should advertise a warranty mileage of around **60,980 miles**!

STEP 17

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1. At least 75,000 miles
2. 60,000 miles or less
3. Between 65,000 and 80,000 miles
4. Warranty for 2% of tires

Let's define our mean $\mu = 70,000$ and standard deviation $\sigma = 4,400$ .
We want to find the probability that a tire lasts at least 75,000 miles.

To do this, we need to calculate the z-score: $z = \frac{x - \mu}{\sigma} = \frac{75,000 - 70,000}{4,400} \approx 1.14$ This tells us how many standard deviations 75,000 is away from the mean.

Now, we look up this z-score in a z-table (or use a calculator) to find the probability of a tire lasting less than 75,000 miles.
Let's say we find this value to be approximately 0.8729.

Since we want the probability of a tire lasting at least 75,000 miles, we subtract the value we just found from 1: $1 - 0.8729 = 0.1271$ So, about 12.71% of tires will last at least 75,000 miles!

We use the same mean and standard deviation as before.
This time, we want the probability of a tire lasting 60,000 miles or less.

Calculate the z-score: $z = \frac{60,000 - 70,000}{4,400} \approx -2.27$

Looking up this z-score in the z-table, we find the probability is approximately 0.0116.
So, only about 1.16% of tires will last 60,000 miles or less.

We need to find the probability of a tire lasting between 65,000 and 80,000 miles.

Calculate the z-scores for both mileages: $z_1 = \frac{65,000 - 70,000}{4,400} \approx -1.14$ $z_2 = \frac{80,000 - 70,000}{4,400} \approx 2.27$

Look up both z-scores in the z-table.
Let's say we find the probabilities to be approximately 0.1271 for $z_1$ and 0.9884 for $z_2$ .

Subtract the smaller probability from the larger one: $0.9884 - 0.1271 = 0.8613$ About 86.13% of tires will last between 65,000 and 80,000 miles!

We want to find the mileage where only 2% of tires will fail.
This means we're looking for a z-score that corresponds to a probability of 0.02.

Looking in the z-table (or using a calculator), we find the z-score closest to 0.02 is approximately -2.05.

Now, we use the z-score formula to find the mileage: $x = \mu + z \cdot \sigma = 70,000 + (-2.05) \cdot 4,400 \approx 60,980$ So, Dunlop should advertise a warranty mileage of around 60,980 miles!