The life of light bulbs is distributed normally. The variance of the lifetime is 225 and the mean lifetime of a bulb is 590 hours. Find the probability of a bulb lasting for at most 607 hour
SD = sq rt of 225 = 15
Z = (score-mean)/SD
Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability related to the Z score.
To find the probability of a bulb lasting for at most 607 hours, we can use the normal distribution.
Step 1: Calculate the standard deviation (σ) using the variance (Var).
σ = √Var = √225 = 15
Step 2: Calculate the standardized value (Z-score) for 607 hours.
Z = (X - μ) / σ
Z = (607 - 590) / 15
Z = 1.13
Step 3: Find the probability using the Z-table.
The Z-table provides the area under the curve to the left of the Z-score. Since we want the probability at most 607 hours, we need to find the area to the left of Z = 1.13.
Looking up the Z-score of 1.13 in the Z-table, we find a value of 0.8708.
Therefore, the probability of a bulb lasting for at most 607 hours is 0.8708 or 87.08%.
To find the probability of a bulb lasting for at most 607 hours, we need to utilize the properties of the normal distribution.
Step 1: Standardize the score
First, we need to standardize the score of 607 hours using the formula:
z = (x - μ) / σ
Where:
- x is the value we want to standardize (607)
- μ is the mean of the distribution (590)
- σ is the standard deviation of the distribution (square root of the variance, which is √225 = 15)
Plugging in the values, we get:
z = (607 - 590) / 15
z = 17 / 15
Step 2: Find the probability
Next, we need to find the probability corresponding to the standardized score using a standard normal distribution table or calculator.
The probability of a bulb lasting for at most 607 hours is equivalent to finding the cumulative probability up to the standardized score z. In other words, we need to find P(Z ≤ z).
Using a standard normal distribution table or calculator, we find that P(Z ≤ 17/15) is approximately 0.8264.
Therefore, the probability of a bulb lasting for at most 607 hours is approximately 0.8264 or 82.64%.