The numbers of hours of life of a torch battery is normally distributed with a mean of 150 hours and standard deviation of 12 hours. In a quality control test, two batteries are chosen at random from a batch. If both batteries have a life less than 120 hours, the batch is rejected. Find the probability that the batch is rejected.
Z = (score-mean)/SD = (120-150)/12
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
The probability of both/all events occurring is found by multiplying the individual probabilities.
I hope this helps.
Well, rejecting the batch just because the batteries didn't live up to their full potential seems a bit harsh, doesn't it? But hey, let's crunch the numbers and find out the probability of this happening.
We have a normally distributed random variable here, representing the hours of life of a torch battery. The mean is 150 hours and the standard deviation is 12 hours. But we're interested in the probability of both batteries having a life less than 120 hours.
To find this probability, we'll have to calculate the probability of a single battery having a life less than 120 hours and then multiply that probability by itself, since we're dealing with two batteries. So let's get to it!
Using the z-score formula, we can calculate the z-score for 120 hours using the formula:
z = (x - μ) / σ
Plugging in the values, we get:
z = (120 - 150) / 12
Calculating this gives us a z-score of -2.5.
Now, we need to find the probability associated with this z-score. Looking up the z-score in a standard normal distribution table (or using a calculator), we find that the probability of a single battery having a life less than 120 hours is approximately 0.0062.
Since we're dealing with two batteries, we need to multiply this probability by itself:
0.0062 * 0.0062 = 0.00003844
So, the probability of both batteries having a life less than 120 hours is approximately 0.00003844.
In other words, it's highly unlikely! Just like finding a unicorn doing a stand-up comedy routine. Good luck with that!
To find the probability that the batch is rejected, we need to calculate the probability that both batteries have a life less than 120 hours.
First, let's calculate the probability that a single battery has a life less than 120 hours. Since the distribution is normal, we can use the Z-score formula:
Z = (x - μ) / σ
where x is the value we want to find the Z-score for (in this case, 120), μ is the mean (150), and σ is the standard deviation (12).
Z = (120 - 150) / 12 = -2.5
Using a standard normal distribution table or calculator, we can find that the probability of a Z-score less than -2.5 is approximately 0.0062.
Now, we need to find the probability that both batteries have a life less than 120 hours. Since the batteries are chosen at random, we can assume that the events are independent. Therefore, we can simply multiply the probabilities:
P(both batteries have a life < 120 hours) = P(battery 1 < 120 hours) * P(battery 2 < 120 hours)
P(both batteries have a life < 120 hours) = 0.0062 * 0.0062 = 0.00003844
Finally, to find the probability that the batch is rejected, we need to subtract the probability that both batteries have a life less than 120 hours from 1 (since if both batteries have a life less than 120 hours, the batch is rejected):
P(batch is rejected) = 1 - P(both batteries have a life < 120 hours) = 1 - 0.00003844 = 0.99996156
Therefore, the probability that the batch is rejected is approximately 0.99996156 or 99.996156%.
To find the probability that the batch is rejected, we need to find the probability that both batteries have a life less than 120 hours.
First, let's calculate the z-score for 120 hours using the formula:
z = (x - μ) / σ
where x is the value we are interested in (120 hours), μ is the mean (150 hours), and σ is the standard deviation (12 hours).
z = (120 - 150) / 12
z = -30 / 12
z = -2.5
Now, we want to find the probability that a battery has a life less than 120 hours, which is equivalent to finding the probability of a z-score less than -2.5. We can use a standard normal distribution table or a calculator to look up the corresponding probability.
Using the standard normal distribution table or a calculator, we find that the probability of a z-score less than -2.5 is approximately 0.0062.
Now, since we have two batteries chosen at random, the probability that both batteries have a life less than 120 hours is calculated by multiplying the probabilities:
P(both batteries have life less than 120) = 0.0062 * 0.0062
P(both batteries have life less than 120) = 0.00003844
Therefore, the probability that the batch is rejected is 0.00003844, or approximately 0.0038%.