Suppose only 75% of all drivers in a certain state regularly wear aseat belt. A random sample of 500 drivers is selected. What is theprobability that
a). Between 360 and 400 (inclusive) of the drivers in the sampleregularly wear a seat belt?
b). Fewer than 400 of those in the sample regularly wear a seatbelt?
Part b)is what I am having a hard time with. This is what I did for part a).
using a Normal approximation:
z = (400.5-500*0.75) /
sqrt(500*0.75*0.25) = 2.63
z = (359.5-500*0.75) /
sqrt(500*0.75*0.25) = -1.60
these numbers correspond to areas of 0.9957 and 0.0548, so the probability is: 0.9957-0.0548 = 0.9409
I need help solving part b).
To solve part b) of the problem, we need to find the probability that fewer than 400 drivers in the sample regularly wear a seat belt.
Using the same normal approximation, we can find the z-score for this case:
z = (399.5 - 500 * 0.75) / sqrt(500 * 0.75 * 0.25) = -2.67
This z-score corresponds to an area of 0.0038 in the standard normal distribution table.
To find the probability of fewer than 400 drivers wearing a seat belt, we need to calculate the cumulative probability up to the z-score of -2.67:
P(Z < -2.67) = 0.0038
Therefore, the probability that fewer than 400 drivers in the sample regularly wear a seat belt is 0.0038.
To solve part b), we need to find the probability that fewer than 400 drivers in the sample regularly wear a seat belt.
We can approach this by using the normal distribution again. However, since we want the probability of fewer than 400, we need to find the cumulative probability up to 400.
Using the normal approximation, we can calculate the z-score for 400:
z = (400 - 500 * 0.75) / sqrt(500 * 0.75 * 0.25)
= (400 - 375) / sqrt(281.25)
= 25 / 16.77
≈ 1.49
To find the probability (P) of fewer than 400 drivers regularly wearing a seat belt, we need to find the cumulative probability up to the z-score of 1.49. This can be done using a standard normal distribution table or a calculator.
P(Z < 1.49) ≈ 0.9319
So, the probability that fewer than 400 of the drivers in the sample regularly wear a seat belt is approximately 0.9319, or 93.19%.
Note that in this calculation, we are assuming that the distribution of the sample proportion is approximately normal due to the Central Limit Theorem.