Passenger on a bus:on the daily run of an express bus,the average number of passengers is 48 . The standard deviation is 3. Assume the variable is normally distributed find the probability that the bus will have
a between 36 and 40
b fewer than 42 passengers
c more than 48 passengers
d between 43 and 47 passengers
To find the probabilities for these different scenarios, we will use the standard normal distribution with a mean of 48 and a standard deviation of 3. Since the variable is assumed to be normally distributed, we can use the Z-score formula to determine the probabilities.
The Z-score formula is given by:
Z = (X - μ) / σ
Where:
Z is the Z-score,
X is the given value,
μ is the mean, and
σ is the standard deviation.
a) To find the probability that the bus will have between 36 and 40 passengers, we need to find the area between the Z-scores corresponding to 36 and 40.
Z1 = (36 - 48) / 3 = -4
Z2 = (40 - 48) / 3 = -2.67
Using a Z-table or calculator, we find the area to the left of Z1 is approximately 0.0003 and the area to the left of Z2 is approximately 0.0038.
Therefore, the probability that the bus will have between 36 and 40 passengers is:
P(36 ≤ X ≤ 40) = P(Z1 ≤ Z ≤ Z2) = P(Z ≤ Z2) - P(Z ≤ Z1) = 0.0038 - 0.0003 = 0.0035.
b) To find the probability of the bus having fewer than 42 passengers, we need to find the area to the left of the Z-score corresponding to 42.
Z = (42 - 48) / 3 = -2
Using the Z-table or calculator, we find the area to the left of Z is approximately 0.0228.
Therefore, the probability that the bus will have fewer than 42 passengers is:
P(X < 42) = P(Z < -2) = 0.0228.
c) To find the probability of the bus having more than 48 passengers, we need to find the area to the right of the Z-score corresponding to 48.
Z = (48 - 48) / 3 = 0
Using the Z-table or calculator, we find the area to the right of Z is approximately 0.5.
Therefore, the probability that the bus will have more than 48 passengers is:
P(X > 48) = P(Z > 0) = 0.5.
d) To find the probability that the bus will have between 43 and 47 passengers, we need to find the area between the Z-scores corresponding to 43 and 47.
Z1 = (43 - 48) / 3 = -1.67
Z2 = (47 - 48) / 3 = -0.33
Using the Z-table or calculator, we find the area to the left of Z1 is approximately 0.0475 and the area to the left of Z2 is approximately 0.3707.
Therefore, the probability that the bus will have between 43 and 47 passengers is:
P(43 ≤ X ≤ 47) = P(Z1 ≤ Z ≤ Z2) = P(Z ≤ Z2) - P(Z ≤ Z1) = 0.3707 - 0.0475 = 0.3232.
So, the probabilities are:
a) 0.0035
b) 0.0228
c) 0.5
d) 0.3232.