The number of hours of life of a certain type of torch battery is normally distributed with mean 150 and standard deviation 12. 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
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability.
For both, multiply probability by itself.
Well, let me calculate that for you. According to the given information, the mean is 150 and the standard deviation is 12. Since we are looking for batteries with a life less than 120 hours, we need to find the probability that a randomly selected battery has a life less than 120 hours.
By consulting my clownish calculations, I can tell you that the probability of a battery having a life less than 120 hours is approximately 0.0918. Now, since we are randomly selecting two batteries, we need to calculate the probability of both batteries having a life less than 120 hours.
To calculate the probability of both batteries having a life less than 120 hours, we multiply the probability of the first battery having a life less than 120 hours by the probability of the second battery having a life less than 120 hours. So the probability is approximately 0.0084.
However, the question asks for the probability that the batch is rejected, which means that both batteries have a life less than 120 hours. So, the probability that the batch is rejected is also approximately 0.0084.
Long story short, the probability that the batch is rejected is like the chances of finding a serious face at a clown convention - extremely low, approximately 0.0084.
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 standardize the values using the formula:
Z = (X - μ) / σ
Where:
X = value
μ = mean
σ = standard deviation
For the first battery:
Z1 = (120 - 150) / 12
= -2.5
For the second battery:
Z2 = (120 - 150) / 12
= -2.5
Now, to find the probability that both batteries have a life less than 120 hours, we need to find the probability that both Z1 and Z2 are less than -2.5.
Since the distribution is normal, we can use a standard normal table or calculator to find this probability.
From the standard normal table, the probability of Z being less than -2.5 is approximately 0.0062.
Since we are calculating the probability of both batteries having a life less than 120 hours, we need to multiply this probability by itself.
P(batch is rejected) = P(Z1 < -2.5) * P(Z2 < -2.5)
= 0.0062 * 0.0062
= 0.00003844
Therefore, the probability that the batch is rejected is approximately 0.00003844 or 0.003844%.
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 standardize the values by using the z-score formula:
z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
For the value 120 hours, the z-score is:
z1 = (120 - 150) / 12 = -2.5
Next, we need to find the probability of selecting a battery with a life less than 120 hours. We can use the standard normal distribution table or a calculator to find this probability. The table gives us the cumulative probability up to a certain z-score.
P(Z < -2.5) = 0.0062 (approximately)
Since we need both batteries to have a life less than 120 hours, we can multiply this probability by itself:
P(both batteries < 120 hours) = 0.0062 * 0.0062 = 0.00003844 (approximately)
This is the probability that a randomly chosen pair of batteries from the batch both have a life less than 120 hours.
To calculate the probability that the batch is rejected, we need to subtract this probability from 1:
P(batch rejected) = 1 - 0.00003844 ≈ 0.99996
Therefore, the probability that the batch is rejected is approximately 0.99996 or 99.996%.