Battery Power Problem. A certain type of thermal battery
for an airplane navigation device backup power has a
mean life of 300 hours with a standard deviation of 15
hours. What proportion of these batteries can be
expected to have lives of 322 hours or less? Assume a normal
distribution of backup power device lives.
To solve this problem, we need to find the proportion of thermal batteries that have lives of 322 hours or less.
First, we need to calculate the z-score. The z-score formula is:
z = (X - μ) / σ
where X is the value we want to find the proportion for (322 hours), μ is the mean (300 hours), and σ is the standard deviation (15 hours).
Let's calculate the z-score:
z = (322 - 300) / 15
z = 22 / 15
z ≈ 1.467
Next, we need to find the proportion associated with the z-score using a standard normal distribution table or calculator.
Let's assume the proportion is represented by the variable P.
P(Z ≤ 1.467) = 0.9276
Therefore, approximately 92.76% (or 0.9276) of these batteries can be expected to have lives of 322 hours or less.
To find the proportion of batteries expected to have lives of 322 hours or less, we need to calculate the probability of a battery having a life less than or equal to 322 hours.
We can use the concept of the standard normal distribution to solve this problem. The standard normal distribution has a mean of 0 and a standard deviation of 1. By standardizing the given data, we can find the corresponding probability using a normal distribution table or a statistical calculator.
To standardize the value of 322 hours, we use the formula:
Z = (X - μ) / σ
Where:
Z is the standard score (also known as the z-score)
X is the given value (322 hours in this case)
μ is the mean (300 hours in this case)
σ is the standard deviation (15 hours in this case)
Plugging in the values:
Z = (322 - 300) / 15
Z = 22 / 15
Z ≈ 1.47
Now, we need to find the proportion of values less than or equal to this standard score. We can refer to a standard normal distribution table to find this probability. Looking up the z-score of 1.47, we find that it corresponds to a cumulative probability of approximately 0.9292.
Therefore, the proportion of batteries expected to have lives of 322 hours or less is approximately 0.9292 or 92.92%.