The lifetime of a brand of flashlight is normally distributed with a
mean of 45 hours and standard deviation of 5 hours. Let X be the lifetime of a randomly selected flashlight battery
of this brand. Then the 𝑃𝑃(40 < X < 50) is not equal to
A) 2𝑃𝑃(45 < X < 50) C) 𝑃𝑃(45 < X < 55)
B) 𝑃𝑃(40 < X < 45) + 𝑃𝑃(45 < X < 50) D) 2𝑃𝑃(45 < X < 50)
The probability that a normally distributed random variable falls within a certain range can be calculated using the standard normal distribution and z-scores.
First, we need to standardize the range of interest by converting the values to z-scores. The formula for calculating z-score is:
z = (x - mean) / standard deviation
For the range (40 < X < 50), we can calculate the z-scores for the lower and upper bounds:
Lower bound: z1 = (40 - 45) / 5 = -1
Upper bound: z2 = (50 - 45) / 5 = 1
Using a standard normal distribution table or a calculator, the probability of a random variable falling between -1 and 1 (inclusive) is approximately 0.6826.
Now let's examine the answer choices:
A) 2P(45 < X < 50):
This suggests calculating the probability of the range (45 < X < 50) and then doubling it. This range has the same z-scores as above (-1 and 1). Doubling the probability would give 2(0.6826) = 1.3652, which is not equal to the probability of the original range.
B) P(40 < X < 45) + P(45 < X < 50):
This suggests calculating the probability of the two individual ranges and summing them. For the range (40 < X < 45), we can calculate the z-scores:
Lower bound: z1 = (40 - 45) / 5 = -1
Upper bound: z2 = (45 - 45) / 5 = 0
Using a standard normal distribution table or a calculator, the probability of a random variable falling between -1 and 0 (inclusive) is approximately 0.3413.
For the range (45 < X < 50), we already calculated the probability as 0.6826.
Summing the probabilities gives 0.3413 + 0.6826 = 1.0239, which is not equal to the probability of the original range.
C) P(45 < X < 55):
This suggests calculating the probability of the range (45 < X < 55). To determine if this probability is equal to the probability of the original range, we need to calculate the z-scores for the upper bound:
Upper bound: z2 = (55 - 45) / 5 = 2
Using a standard normal distribution table or a calculator, the probability of a random variable falling between -1 and 2 (inclusive) is approximately 0.8186. This is not equal to the probability of the original range.
D) 2P(45 < X < 50):
This suggests calculating the probability of the range (45 < X < 50) and then doubling it. This range has the same z-scores as above (-1 and 1). Doubling the probability would give 2(0.6826) = 1.3652, which is not equal to the probability of the original range.
Therefore, the correct answer is none of the above.