Lifetime of a tire Assume the random variable L in Example 2f ...

Question

Lifetime of a tire Assume the random variable L in Example 2f is normally distributed with mean μ = 22,000 miles and σ = 4,000 miles.

a. In a batch of 4000 tires, how many can be expected to last for at least 18,000 miles?

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A manufacturer of LED light bulbs guarantees that each bulb will emit light for at least 15,000 hours. However, actual lifetimes vary. Let the lifetime T in hours of an LED light bulb be modeled by a continuous random variable with a normal distribution mean mu is equal to 13,000 hours and standard deviation sigma is equal to 3000 hours. In a shipment of 5000 bulbs, how many would you expect to reach at least 15,000 hours of operation? OK. So it appears for this particular prong we're asked to take all the information that is provided to us and we're asked to determine within a shipment of 5000 LED light bulbs, how many of these light bulbs would we expect to reach at least 15,000 hours of operation. So we're trying to determine how many, so we're gonna need to make sure our final answers all in a whole number. Not a half number, not a negative number, it has to be a whole value, so it's real. So we're trying to determine how many light bulbs in this batch of 5000 bulbs will reach at least 15,000 hours of operation. So that is what we're ultimately trying to solve for. So now we know what we're ultimately trying to solve for. Basically, what we're gonna need to do in order to solve this problem is we will need to calculate the probability that the normally distributed, once again, this normally distributed bulb lifetime T will exceed 15,000 hours, and we will do this by recalling and using the integral of normal probability density function. Then we will need to multiply this probability by 5000 in order to find the expected number of bulbs that is meeting that specific lifetime value. And we need to recall and note that the lifetime T value in units of hours will be normally distributed with mu being equal to 13,000, and once again, mu indicates the normal distribution mean. And then we have our standard deviation sigma, which will be equal to, once again, 3000 hours. Fantastic. So with that in mind, we need to recall a note that the normal distribution is defined by the following probability density function, which will be written as. F of x is equal to 1 divided by sigma multiplied by the square root of 2 pi multiplied by e to the power of 1/2 multiplied by parentheses of x minus mu divided by sigma to the power of 2. We also need to recall a note based on our prior knowledge that the normal distribution is symmetric, about the mean mu, and the total area under the curve will be 1. And to help us visualize this prong, let's create a quick sketch of the curve for this normal distribution curve, this normal distribution graph. And when we do our normal distribution curve should look as follows. So as we can see, we have our curve represented in blue, and we have a red dotted line going up to the peak of our curve which denotes our mean. And as we can see, we have our mean mu, which is 13,000, and then we can see that our curve starts to drop off, starts to decrease as it heads towards 15,000. And that's our x axis, which is the hours of use. This is our lifetime value that we're trying to use. So it's basically the number of hours used. So, Now that we have a good visualization of what our curve looks like for this, once again, we are looking at a, uh, you know, to reiterate, this is a, so we're trying to find the lifetime variable, and we are ultimately focusing using this normal distribution curve to do that. So, Looking now to the, to our next step that we need to consider is we need to note that half of the total area will lie to the left and half to the right, which will mean we can go ahead and write that P, which P represents probability. So PT is less than 13,000 will be equal to PT is greater than 13,000, which will be equal to 1/2. And the following integral that we need to focus on now will represent the probability that T will lie between 13,000, which is the mean, and 15,000, meaning we could write the integral evaluated from 13,000 to 15,000 of 1 divided by 3000 multiplied by the square root of 2 pi multiplied by e to the power of -1/2. Multiplied by parentheses of x minus 13,000. Divided by 3000 to the power of 2 DX, fantastic, and since the total area of the right of 13,000 is 1/2. And the area between 13,000 and 15,000 is the integral above, that will mean, as a result, the area to the right of 15,000 will be written as P of T is greater than or equal to 15,000, which will be equal to 1/2 minus the integral evaluated from 13,000 to 15,000 of 1 divided by. 3000 multiplied by the square root of 2 pi multiplied by e to the power of negative 1/2 multiplied by parentheses of x minus 13,000 divided by 3000. To the power of 2, so this will be to the power of 2. Fantastic. We are on a roll here. So, and don't forget our DX, and when we plug this expression into our calculator, we will find that it's approximately equal to 0.2524925375. And we now need to multiply this probability by the shipment size of 5000 LED light bulbs, and the expected number of bulbs lasting at at least 15,000 hours will be. 5000 multiplied by 0.2524925375 will be approximately equal to rounding to the nearest whole number 1260. And that is our final answer. Hooray, we did it. So therefore, without a doubt, approximately 1,262 light bulbs will reach at least 15,000 hours of operation. So looking at the prom itself, we can safely say once again that our final answer without a doubt will be. For our number of light bulbs will be 1,262, and that's it. Thank you so much for watching, hopefully that helped, and I can't wait to see you in the next video. Bye.

Lifetime of a tire Assume the random variable L in Example 2f is normally distributed with mean μ = 22,000 miles and σ = 4,000 miles.a. In a batch of 4000 tires, how many can be expected to last for at least 18,000 miles?

Watch next

Lifetime of a tire Assume the random variable L in Example 2f is normally distributed with mean μ = 22,000 miles and σ = 4,000 miles.
a. In a batch of 4000 tires, how many can be expected to last for at least 18,000 miles?