Heights of 10 year olds. Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches.
(a) The height requirement for Batman the Ride at Six Flags Magic Mountain is 54 inches. What percent of 10 year olds cannot go on this ride?
(b) Suppose there are four 10 year olds. What is the chance that at least two of them will be able to ride Batman the Ride?
(c) Suppose you work at the park to help them better understand their customers’ demographics, and you are counting people as they enter the park. What is the chance that the first 10 year old you see who can ride Batman the Ride is the 3rd 10 year old who enters the park?
(d) What is the chance that the fifth 10 year old you see who can ride Batman the Ride is the 12th 10 year old who enters the park?
(a) To find the percent of 10-year-olds who cannot go on the ride, we need to calculate the z-score for a height of 54 inches using the formula:
z = (x - μ) / σ
where x is the height, μ is the mean, and σ is the standard deviation.
For this case, x = 54 inches, μ = 55 inches, and σ = 6 inches.
z = (54 - 55) / 6 = -1/6
Using the Z-table or a calculator, we can find the probability corresponding to a z-score of -1/6, which is approximately 0.4192.
Therefore, approximately 41.92% of 10-year-olds cannot go on the Batman the Ride.
(b) To find the chance that at least two out of four 10-year-olds will be able to ride the Batman the Ride, we need to find the probability of two, three, or four of them being able to ride.
Let's calculate the probability for each case:
For two 10-year-olds being able to ride:
P(2 10-year-olds) = P(1st 10-year-old can ride) * P(2nd 10-year-old can ride) * P(3rd 10-year-old cannot ride) * P(4th 10-year-old cannot ride)
P(2 10-year-olds) = (1 - 0.4192) * (1 - 0.4192) * 0.4192 * 0.4192
For three 10-year-olds being able to ride:
P(3 10-year-olds) = P(1st 10-year-old can ride) * P(2nd 10-year-old can ride) * P(3rd 10-year-old can ride) * P(4th 10-year-old cannot ride)
P(3 10-year-olds) = (1 - 0.4192) * (1 - 0.4192) * (1 - 0.4192) * 0.4192
For four 10-year-olds being able to ride:
P(4 10-year-olds) = P(1st 10-year-old can ride) * P(2nd 10-year-old can ride) * P(3rd 10-year-old can ride) * P(4th 10-year-old can ride)
P(4 10-year-olds) = (1 - 0.4192) * (1 - 0.4192) * (1 - 0.4192) * (1 - 0.4192)
Now, let's calculate the chances:
P(at least 2 of 4 10-year-olds can ride) = P(2 10-year-olds) + P(3 10-year-olds) + P(4 10-year-olds)
P(at least 2 of 4 10-year-olds can ride) = [(1 - 0.4192) * (1 - 0.4192) * 0.4192 * 0.4192] + [(1 - 0.4192) * (1 - 0.4192) * (1 - 0.4192) * 0.4192] + [(1 - 0.4192) * (1 - 0.4192) * (1 - 0.4192) * (1 - 0.4192)]
Simplifying the expression gives us the final probability.
(c) To find the chance that the first 10-year-old who can ride Batman the Ride is the 3rd 10-year-old to enter the park, we need to consider the probability that the first two 10-year-olds cannot ride and the third one can.
P(first 10-year-old who can ride is the 3rd) = P(1st 10-year-old cannot ride) * P(2nd 10-year-old cannot ride) * P(3rd 10-year-old can ride)
P(first 10-year-old who can ride is the 3rd) = 0.4192 * 0.4192 * (1 - 0.4192)
(d) Similarly, to find the chance that the fifth 10-year-old who can ride Batman the Ride is the 12th 10-year-old to enter the park, we need to consider the probability that the first 11 10-year-olds cannot ride and the 12th one can.
P(fifth 10-year-old who can ride is the 12th) = P(1st 10-year-old cannot ride) * P(2nd 10-year-old cannot ride) * P(3rd 10-year-old cannot ride) * P(4th 10-year-old cannot ride) * P(5th 10-year-old can ride)
P(fifth 10-year-old who can ride is the 12th) = 0.4192 * 0.4192 * 0.4192 * 0.4192 * (1 - 0.4192)
To answer the questions, we need to use the properties of the normal distribution. The normal distribution is characterized by its mean and standard deviation. In this case, the mean height of 10 year olds is 55 inches, and the standard deviation is 6 inches.
(a) To find the percent of 10 year olds who cannot go on Batman the Ride, we need to calculate the percentage of 10 year olds who are shorter than 54 inches. This can be done by finding the area under the normal curve to the left of 54 inches.
To find this, we can use the formula for the standard normal distribution:
Z = (x - μ) / σ
Where:
Z = the z-score (standardized value)
x = the value we want to convert to a z-score (54)
μ = the mean (55)
σ = the standard deviation (6)
Using these values, we can find the z-score:
Z = (54 - 55) / 6 = -1/6 = -0.1667
Now, we need to find the area to the left of the z-score using a standard normal distribution table or a calculator. The area to the left of -0.1667 is approximately 0.4332.
Therefore, the percent of 10 year olds who cannot go on Batman the Ride is approximately 43.32%.
(b) To find the chance that at least two out of four 10 year olds can ride Batman the Ride, we will calculate the complementary probability of the event where less than two 10 year olds can ride.
The probability that a 10 year old can ride Batman the Ride is the same as the result from part (a), which is 1 - 0.4332 = 0.5668.
The probability that none of the four 10 year olds can ride is (0.5668)^4 = 0.1025. This means that 10.25% of the time, none of the four 10 year olds can ride.
Therefore, the probability that at least two out of four 10 year olds can ride Batman the Ride is 1 - 0.1025 = 0.8975, or approximately 89.75%.
(c) To find the chance that the first 10 year old you see who can ride Batman the Ride is the third 10 year old who enters the park, we need to calculate the probability that the first two 10 year olds cannot ride and the third one can.
The probability that a 10 year old cannot ride is the same as the result from part (a), which is 0.4332.
The probability that the first two cannot ride is (1 - 0.4332)^2 = 0.3071.
The probability that the third one can ride is 1 - 0.4332 = 0.5668.
Therefore, the chance that the first 10 year old you see who can ride Batman the Ride is the third 10 year old who enters the park is 0.3071 * 0.5668 = 0.1743, or approximately 17.43%.
(d) Similarly, to find the chance that the fifth 10 year old you see who can ride Batman the Ride is the twelfth 10 year old who enters the park, we need to calculate the probability that the first eleven 10 year olds cannot ride and the twelfth one can.
The probability that a 10 year old cannot ride is the same as the result from part (a), which is 0.4332.
The probability that the first eleven cannot ride is (1 - 0.4332)^11 = 0.0277.
The probability that the twelfth one can ride is 1 - 0.4332 = 0.5668.
Therefore, the chance that the fifth 10 year old you see who can ride Batman the Ride is the twelfth 10 year old who enters the park is 0.0277 * 0.5668 = 0.0157, or approximately 1.57%.