The lifetimes of batteries produced by a firm are known to be normally distributed with a mean of 100 hours and a standard deviation of 10 hours. What is the probability a battery will last between 110 and 120 hours?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z scores.
The lifetime (in years) of 9 automobile batteries of a certain brand are: 2.4, 1.9, 2.0, 2.1, 1.8, 2.3, 2.1, 2.3, 1.7. At alpha O.05 level of significance, test if the mean battery life is less than 25 years.
State the null and alternative hypothesis.
State your conclusion for your hypothesis
Give a conclusion for the confidence interval
To find the probability that a battery will last between 110 and 120 hours, we need to calculate the area under the normal curve between these two values.
Step 1: Standardize the values
First, we need to standardize the values of 110 and 120. Standardizing means converting the values to "z-scores" using the formula:
z = (x - μ) / σ
where z is the z-score, x is the value, μ is the mean, and σ is the standard deviation.
For 110 hours:
z1 = (110 - 100) / 10 = 1
For 120 hours:
z2 = (120 - 100) / 10 = 2
Step 2: Find the probabilities
Once we have calculated the z-scores, we can use a standard normal distribution table or a calculator with a standard normal distribution function to find the probabilities associated with these z-scores.
Using a standard normal distribution table, we can find the corresponding probabilities:
For z1 = 1, the corresponding probability is 0.8413.
For z2 = 2, the corresponding probability is 0.9772.
Step 3: Calculate the probability between the two values
To find the probability between 110 and 120 hours, we subtract the probability associated with the z-score of 110 from the probability associated with the z-score of 120:
P(110 ≤ X ≤ 120) = P(Z ≤ 2) - P(Z ≤ 1)
= 0.9772 - 0.8413
= 0.1359
Therefore, the probability that a battery will last between 110 and 120 hours is approximately 0.1359, or 13.59%.