(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)
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?
1 answer