Question

A certain brand of automobile tire has a mean life span of 35,000 miles, with a standard deviation of 2250 miles. Assume the life spans of the tires have a bell-shaped distribution.
(a) The life spans of three randomly selected tires are 34,000 miles, 37,000 miles, and 30,000 miles. Find the z-score that corresponds to each life span. Determine whether any of these life spans are unusual.
(b) The life spans of three randomly selected tires are 30,500 miles, 37,250 miles, and 35,000 miles. Using the Empirical Rule, find the percentile that corresponds to each life span.

Answer 1

Answer:

Step-by-step explanation:

From the information given:

mean life span of a brand of automobile = 35,000

standard deviation of a brand of automobile = 2250 miles.

the z-score that corresponds to each life span are as follows.

the standard z- score formula is:

$z = \dfrac{x - \mu}{\sigma}$

For x = 34000

$z = \dfrac{34000 - 35000}{2250}$

$z = \dfrac{-1000}{2250}$

z = −0.4444

For x = 37000

$z = \dfrac{37000 - 35000}{2250}$

$z = \dfrac{2000}{2250}$

z = 0.8889

For x = 3000

$z = \dfrac{30000 - 35000}{2250}$

$z = \dfrac{-5000}{2250}$

z = -2.222

From the above z- score that corresponds to their life span; it is glaring  that the tire with the life span 30,000 miles has an unusually short life span.

For x = 30,500

$z = \dfrac{30500 - 35000}{2250}$

$z = \dfrac{-4500}{2250}$

z = -2

P(z) = P(-2)

Using excel function (=NORMDIST -2)

P(z) = 0.022750132

P(z) = 2.28th percentile

For x =  37250

$z = \dfrac{37250 - 35000}{2250}$

$z = \dfrac{2250}{2250}$

z = 1

Using excel function (=NORMDIST 1)

P(z) = 0.841344746

P(z) = 84.14th percentile

For x = 35000

$z = \dfrac{35000- 35000}{2250}$

$z = \dfrac{0}{2250}$

z = 0

Using excel function (=NORMDIST 0)

P(z) = 0.5

P(z) = 50th percentile

Answer 2

a.  The z score of each life span should be -0.4444, 0.889, and 2.2222.

b.  The percentile of each life span should be 0.0228, 0.8413 and  0.5000.

Given that,

• mean life span of 35,000 miles, with a standard deviation of 2250 miles.

The calculation is as follows:

(a)

The z score should be

$Z1 = \frac{34000-35000}{2250} = -0.4444\\\\Z2 = \frac{37000-35000}{2250} = 0.8889\\\\Z3 = \frac{30000-35000}{2250} = -2.2222$

The tire with life span of 30000 miles would be considered unusual

(b)

The percentile should be

$Z1 = \frac{30500-35000}{2250} = -2$

p(Z1 < -2) = NORMSDIST(-2) = 0.0228

$Z2 = \frac{37250-35000}{2250} = 1$

p(Z2 < 1) = NORMSDIST(1) = 0.8413

$Z3 = \frac{35000-35000}{2250} = 0$

p(Z3 < 0) = NORMSDIST(0) = 0.5000

Find out more information about standard deviation here:

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