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
Between 36 and 40 passengers
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z scores.
ette
To find the probability that the bus will have between 36 and 40 passengers, we will use the standard normal distribution.
Step 1: Convert the individual values to z-scores using the formula:
z = (x - μ) / σ
Where:
x = individual value
μ = mean
σ = standard deviation
For 36 passengers:
z1 = (36 - 48) / 3 = -4. Therefore, z1 = -4.
For 40 passengers:
z2 = (40 - 48) / 3 = -2.67. Therefore, z2 ≈ -2.67.
Step 2: Use the z-scores to find the corresponding probabilities using a standard normal distribution table or calculator.
P(36 < X < 40) = P(-4 < Z < -2.67)
Approximately, P(-4 < Z < -2.67) = P(Z < -2.67) - P(Z < -4)
Look up the z-scores in the standard normal distribution table to find the corresponding probabilities.
P(Z < -2.67) ≈ 0.0034
P(Z < -4) ≈ 0.00003
Step 3: Calculate the final probability by subtracting P(Z < -4) from P(Z < -2.67).
P(-4 < Z < -2.67) ≈ 0.0034 - 0.00003
So, the probability that the express bus will have between 36 and 40 passengers is approximately 0.00337.
To find the probability that the bus will have between 36 and 40 passengers, we need to use the concept of the standard normal distribution.
Step 1: Convert the given values to z-scores.
To find the z-score for each value, we can use the formula: (x - μ) / σ
where x is the given value, μ is the mean, and σ is the standard deviation.
For 36 passengers:
z-score = (36 - 48) / 3 = -4
For 40 passengers:
z-score = (40 - 48) / 3 = -2.67
Step 2: Find the probability associated with each z-score using the standard normal distribution table.
The standard normal distribution table provides the area under the curve to the left of a given z-score.
Using the table, we find that the probability associated with a z-score of -4 is approximately 0.00003168.
Using the table, we find that the probability associated with a z-score of -2.67 is approximately 0.00383.
Step 3: Calculate the probability between the two z-scores.
To find the probability between two z-scores, we subtract the probability associated with the lower z-score from the probability associated with the higher z-score.
Probability between 36 and 40 passengers = Probability (z-score ≤ -2.67) - Probability (z-score ≤ -4)
= 0.00383 - 0.00003168
= 0.00379832
Therefore, the probability that the bus will have between 36 and 40 passengers is approximately 0.0038, or 0.38%.