The average time a visitor spends at the Renzie Park Art Exhibit is 62 minutes. The standard deviation is 12 minutes. If a visitor is selected at random, find the probability that he or she will spend:
A)At least 82 minutes at the exhibit
B)At most 50 minutes at the exhibit
The average time a person spends at Barefoot Landing Seaquarium is 96 minutes. The standard deviation is 17 minutes. Assume the variable is normally distributed. If a visitor is selected at random, find the probability that he or she will spend:
A)At least 120 minutes at the exhibit
B)At most 80 minutes at the exhibit
To solve these problems, we will use the z-score formula and the standard normal distribution table. The z-score formula is given by:
z = (x - μ) / σ
Where:
- x represents the value we want to find the probability for.
- μ represents the mean of the distribution.
- σ represents the standard deviation of the distribution.
For all the problems, we need to find the probability under the curve to the left or right of the given value. To do this, we will use the standard normal distribution table, which gives the area under the curve to the left of a specific z-score value.
Let's solve these problems step-by-step:
1. Renzie Park Art Exhibit:
- Mean (μ) = 62 minutes
- Standard Deviation (σ) = 12 minutes
A) To find the probability that a visitor spends at least 82 minutes at the exhibit, we need to find the area under the curve to the right of 82 minutes.
z = (82 - 62) / 12 = 20 / 12 = 1.67
Using the standard normal distribution table, we find that the area to the left of 1.67 is 0.9525. To find the area to the right, we subtract this value from 1:
Probability = 1 - 0.9525 = 0.0475 (or 4.75%)
B) To find the probability that a visitor spends at most 50 minutes at the exhibit, we need to find the area under the curve to the left of 50 minutes.
z = (50 - 62) / 12 = -12 / 12 = -1
Using the standard normal distribution table, we find that the area to the left of -1 is 0.1587.
Probability = 0.1587 (or 15.87%)
2. Barefoot Landing Seaquarium:
- Mean (μ) = 96 minutes
- Standard Deviation (σ) = 17 minutes
A) To find the probability that a visitor spends at least 120 minutes at the exhibit, we need to find the area under the curve to the right of 120 minutes.
z = (120 - 96) / 17 = 24 / 17 = 1.41
Using the standard normal distribution table, we find that the area to the left of 1.41 is 0.9207. To find the area to the right, we subtract this value from 1:
Probability = 1 - 0.9207 = 0.0793 (or 7.93%)
B) To find the probability that a visitor spends at most 80 minutes at the exhibit, we need to find the area under the curve to the left of 80 minutes.
z = (80 - 96) / 17 = -16 / 17 = -0.94
Using the standard normal distribution table, we find that the area to the left of -0.94 is 0.1736.
Probability = 0.1736 (or 17.36%)
To solve these probability questions, we will use the concept of the standard normal distribution.
The standard normal distribution is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. By using standardization, we can convert any normal distribution to the standard normal distribution.
To find the probability for each question, we need to convert the given values to z-scores and then use the z-table (also known as the standard normal distribution table) to find probabilities.
Let's start with the first question:
A) To find the probability that a visitor will spend at least 82 minutes at the exhibit, we need to calculate the z-score for 82 minutes and find the corresponding probability.
To calculate the z-score, we use the formula:
z = (x - μ) / σ
where x is the value we want to calculate the probability for, μ is the mean, and σ is the standard deviation.
In this case, x = 82 minutes, μ = 62 minutes, and σ = 12 minutes.
z = (82 - 62) / 12
z = 20 / 12
z ≈ 1.67
Now we can use the z-table to find the probability corresponding to a z-score of 1.67.
Looking up the value 1.67 in the z-table, we find that the probability is approximately 0.9525.
However, we are looking for the probability of spending at least 82 minutes, which means we need to find the area to the right of the z-score.
Since the total area under the normal curve is 1, the probability of spending at least 82 minutes is 1 - 0.9525 = 0.0475.
Therefore, the probability that a visitor will spend at least 82 minutes at the Renzie Park Art Exhibit is approximately 0.0475.
B) To find the probability that a visitor will spend at most 50 minutes at the exhibit, we follow a similar process.
To calculate the z-score, we use the formula:
z = (x - μ) / σ
In this case, x = 50 minutes, μ = 62 minutes, and σ = 12 minutes.
z = (50 - 62) / 12
z = -12 / 12
z = -1
Using the z-table, we find that the probability corresponding to a z-score of -1 is approximately 0.1587.
Since we are looking for the probability of spending at most 50 minutes, which means we need to find the area to the left of the z-score, the probability is 0.1587.
Therefore, the probability that a visitor will spend at most 50 minutes at the Renzie Park Art Exhibit is approximately 0.1587.
Let's move on to the second set of questions about the Barefoot Landing Seaquarium:
A) To find the probability that a visitor will spend at least 120 minutes at the exhibit, we need to calculate the z-score for 120 minutes and find the corresponding probability.
To calculate the z-score, we use the formula:
z = (x - μ) / σ
In this case, x = 120 minutes, μ = 96 minutes, and σ = 17 minutes.
z = (120 - 96) / 17
z = 24 / 17
z ≈ 1.41
Using the z-table, we find that the probability corresponding to a z-score of 1.41 is approximately 0.9207.
Since we are looking for the probability of spending at least 120 minutes, which means we need to find the area to the right of the z-score, the probability is 0.9207.
Therefore, the probability that a visitor will spend at least 120 minutes at the Barefoot Landing Seaquarium is approximately 0.9207.
B) To find the probability that a visitor will spend at most 80 minutes at the exhibit, we follow a similar process.
To calculate the z-score, we use the formula:
z = (x - μ) / σ
In this case, x = 80 minutes, μ = 96 minutes, and σ = 17 minutes.
z = (80 - 96) / 17
z = -16 / 17
z ≈ -0.94
Using the z-table, we find that the probability corresponding to a z-score of -0.94 is approximately 0.1736.
Since we are looking for the probability of spending at most 80 minutes, which means we need to find the area to the left of the z-score, the probability is 0.1736.
Therefore, the probability that a visitor will spend at most 80 minutes at the Barefoot Landing Seaquarium is approximately 0.1736.