A luxury passenger liner has 500 passengers on board whose ages are normally distributed around a mean of 60 years with a standard deviation of 12 years. What is the probability of passengers less than 55 years old?
To find the probability of passengers less than 55 years old, we need to calculate the z-score for an age of 55 years using the formula:
z = (x - μ) / σ
Where:
x = 55 (the age we are interested in)
μ = 60 (mean)
σ = 12 (standard deviation)
Plugging in the values:
z = (55 - 60) / 12
z = -5 / 12
z = -0.4167
Next, we find the corresponding probability using a standard normal distribution table or a calculator. The probability of a z-score of -0.4167 or less can be found to be approximately 0.3372.
Therefore, the probability of passengers less than 55 years old is approximately 0.3372 or 33.72%.
To find the probability of passengers being less than 55 years old, we need to calculate the z-score first. The z-score measures the number of standard deviations an individual value is from the mean.
Step 1: Calculate the z-score.
The formula to calculate the z-score is:
z = (x - μ) / σ
Where:
x = The value we want to find the probability for (in this case, 55 years old)
μ = The mean of the distribution (60 years old)
σ = The standard deviation of the distribution (12 years)
Plugging in the values, we get:
z = (55 - 60) / 12
z = -0.41667
Step 2: Look up the z-score in the standard normal distribution table.
The standard normal distribution table provides the probability associated with a given z-score. In this case, we want to find the probability of passengers being less than 55 years old, so we need to find the probability to the left of the z-score (-0.41667).
Looking up the z-score in the table, we find that the closest value is 0.4162 (which corresponds to a cumulative probability of 0.6615).
Step 3: Calculate the final probability.
Since our z-score is negative, the cumulative probability of passengers being less than 55 years old can be calculated as:
P(x < 55) = 0.5 - 0.6615
P(x < 55) = 0.1585
Therefore, the probability of passengers being less than 55 years old is approximately 0.1585 or 15.85%.
To find the probability of passengers on the luxury liner being less than 55 years old, we can use the properties of the normal distribution.
Step 1: Standardize the value
First, we need to standardize the value 55 using the formula for standardization:
z = (x - μ) / σ
Where:
- x is the value we want to standardize (55 in this case)
- μ is the mean of the distribution (60 in this case)
- σ is the standard deviation of the distribution (12 in this case)
Plugging in the values, we get:
z = (55 - 60) / 12
z = -0.4167 (rounded to four decimal places)
Step 2: Find the probability
Once we have the standardized value, we can find the corresponding probability using a z-table or a statistical software.
Looking up the z-value (-0.4167) in a standard normal distribution table, we find that the cumulative probability is 0.3413 or approximately 0.34 (rounded to two decimal places).
Therefore, the probability of passengers on the luxury liner being less than 55 years old is 0.34 or 34%.