Given : A manufacturer of radial tires for automobiles has extensive data to support the fact that the lifetime of their tires follows a normal distribution with

$\mu=42,100\text{ miles}$

$\sigma=2,510\text{ miles}$

Let x be the random variable that represents the lifetime of the tires .

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

For x= 44,500 miles

$z=\dfrac{44500-42100}{2510}\approx0.96$

For x= 48,000 miles

$z=\dfrac{48000-42100}{2510}\approx2.35$

Using the standard normal distribution table , we have

The p-value : $P(44500$

$P(z$

Hence, the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles = 0.1591

3 0

2 years ago

64.3 - 3 x 2 by the power of 3 divided by two

kolbaska11 [484]

If you meant:

64.3-3*(2³)/2

That would be 52.3 :)

8 0

2 years ago

Read 2 more answers

$1,240 at 8% compounded<br> annually for 2 years

saw5 [17]

The answer is $1,446.33. 1,240 x 8% = $99.20 $1,240 + $99.20 = $1,339.20. $1,339.20 x 8% = $107.13 $1,339.20 + $107.13 = $1,446.33. Hope I could help! :D

4 0

2 years ago