1. The age of professional racecar drivers is normally distributed with a mean of 30 and a standard deviation of 6. One driver is randomly selected. What is the probability that this driver is between 31 and 36 years old?
a. .023
b. .878
c. .275
d. not enough information to answer
2. The age of professional racecar drivers is normally distributed with a mean of 30 and a standard deviation of 6. One driver is randomly selected. What is the probability that this driver is less than 18 years old?
a. .023
b. .878
c. .275
d. not enough information to answer
d. not enough information to answer
1. Well, the age of professional racecar drivers is like going around a track - it follows a normal distribution! So let's calculate the probability of this driver being between 31 and 36 years old. We need to find the area under the bell curve, also known as the cumulative distribution function (CDF). The z-scores for 31 and 36 are (31-30)/6 = 0.1667 and (36-30)/6 = 1.0, respectively. Using a standard normal distribution table or a calculator, we find that the area to the left of z = 0.1667 is 0.5656 and the area to the left of z = 1.0 is 0.8413. So, to find the probability this driver is between 31 and 36 years old, we calculate 0.8413 - 0.5656 = 0.2757. Round that to 0.275, and we come to the hilarious answer of... c. 0.275!
2. Ah, it seems we have another driver who is just revving their engine, eager to get on the track! We want to find the probability that this driver is less than 18 years old. Since our normal distribution has a mean of 30 and a standard deviation of 6, we can calculate the z-score for 18 using the formula (18-30)/6 = -2. Calculate, calculate... and we find that the area to the left of z = -2 is 0.0228. So, the probability this driver is less than 18 years old is approximately 0.0228. Roll on, funny answer a. 0.023!
1. To find the probability that the driver is between 31 and 36 years old, we need to calculate the z-scores for both ages and then find the area under the normal distribution curve between these two z-scores.
First, let's calculate the z-score for the lower age of 31:
z1 = (31 - 30) / 6 = 0.1667
Next, let's calculate the z-score for the upper age of 36:
z2 = (36 - 30) / 6 = 1.0000
To find the probability between these two z-scores, we can use a standard normal distribution table or a calculator. Using the table, we can find the corresponding probabilities for z1 and z2:
P(z < 0.1667) = 0.5662
P(z < 1.0000) = 0.8413
To find the probability between these two values, we subtract the lower probability from the higher probability:
P(31 < x < 36) = P(z < 1.0000) - P(z < 0.1667) = 0.8413 - 0.5662 = 0.2751
Therefore, the probability that the driver is between 31 and 36 years old is approximately 0.275.
The correct answer is c. 0.275.
2. To find the probability that the driver is less than 18 years old, we need to calculate the z-score for 18 and find the area under the normal distribution curve to the left of this z-score.
The z-score for 18 is:
z = (18 - 30) / 6 = -2
To find the probability for z < -2, we can use a standard normal distribution table or a calculator. Using the table, we can find the corresponding probability:
P(z < -2) = 0.0228
Therefore, the probability that the driver is less than 18 years old is approximately 0.0228.
The correct answer is a. 0.023.
To find the probability in both of these scenarios, we need to use the concept of the standard normal distribution. The standard normal distribution has a mean of 0 and a standard deviation of 1.
To calculate probability in these scenarios, we can use z-scores. The z-score measures how many standard deviations an individual value is from the mean.
1. To find the probability that a driver is between 31 and 36 years old, we need to transform these values into z-scores.
The z-score formula is: z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.
For 31 years old: z1 = (31 - 30) / 6 = 1/6 = 0.167
For 36 years old: z2 = (36 - 30) / 6 = 6/6 = 1
Next, we need to find the probability between these two z-scores. This can be done using a standard normal distribution table or a calculator.
Using a standard normal distribution table, we can look up the area corresponding to each z-score and subtract the smaller area from the larger area.
The area corresponding to z1 = 0.167 is 0.5675 (from the table).
The area corresponding to z2 = 1 is 0.8413 (from the table).
Therefore, the probability that a driver is between 31 and 36 years old is: P(31 ≤ x ≤ 36) = 0.8413 - 0.5675 = 0.2738.
Choosing the closest answer from the given options, the probability is approximately 0.275 (option c).
2. For this scenario, we need to find the probability that a driver is less than 18 years old.
First, we calculate the z-score for 18 years old: z = (18 - 30) / 6 = -12 / 6 = -2.
Using a standard normal distribution table, we can find the area corresponding to z = -2, which is 0.0228.
Therefore, the probability that a driver is less than 18 years old is approximately 0.0228.
Choosing the closest answer from the given options, the probability is approximately 0.023 (option a).