Tires of a particular brand have a lifetime mean of 29,000 miles and a standard deviation of 3,000 miles. (20 points) a. It can be guaranteed that 75% of the lifetimes of tires of this brand will be in what interval? B. Using the empirical rule, it can be estimated that approximately 68% of the lifetimes of tires of this brand will be in what interval?

a. The interval: 26,978 miles and 32,022 miles

b. The interval: 26,000 miles and 32,000 miles

Step-by-step explanation:

a. It can be guaranteed that 75% of the lifetimes of tires of this brand will be in what interval?

0.674

We solve question a, using z score formula.

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 29,000 miles

σ is the population standard deviation = 3,000 miles

We are asked to find the interval = x

z = z score of 75th and 25th percentile = ±0.674

Hence:

For z = -0.674

-0.674 = x - 29,000/3,000

Cross Multiply

-0.674 × 3,000 = x - 29,000

-2022 = x - 29,000

x = - 2022 + 29,000

x = 26,978 miles

For z = 0.674

0.674 = x - 29,000/3,000

Cross Multiply

0.674 × 3,000 = x - 29,000

2022 = x - 29,000

x = 2022 + 29,000

x = 31,022 miles

The interval: 26,978 miles and 32,022 miles

b. Using the empirical rule, it can be estimated that approximately 68% of the lifetimes of tires of this brand will be in what interval?

68% of data falls within 1 standard deviations from the mean - between μ – σ and μ + σ .

Mean (μ) = 29,000 miles

Standard deviation (σ) = 3,000 miles.

Hence the interval is calculated as:

μ – σ

= 29,000 - 3,000

= 26,000 miles

μ + σ

= 29,000 + 3,000

= 32,000 miles

The interval: 26,000 miles and 32,000 miles

where are the choices?

the photograph is photographer and- in the process- ourselves.

step-by-step explanation: