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Monsoon Rains. The summer monsoon rains in India follow approximately a Normal distribution with mean 852 millimeters \((\mathrm{mm}\) ) of rainfall and standard deviation \(82 \mathrm{~mm}\). a. In the drought year \(1987,697 \mathrm{~mm}\) of rain fell. In what percentage of all years will India have \(697 \mathrm{~mm}\) or less of monsoon rain? b. "Normal rainfall" means within \(20 \%\) of the long-term average, or between \(682 \mathrm{~mm}\) and \(1022 \mathrm{~mm}\). In what percentage of all years is the rainfall normal?

Short Answer

Expert verified
a) 2.94%; b) 96.16%

Step by step solution

01

Calculate the Z-score for 697 mm

To find the percentage of years with 697 mm or less rainfall, we first need to calculate the Z-score. The formula for the Z-score is \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the given value (697 mm), \( \mu \) is the mean (852 mm), and \( \sigma \) is the standard deviation (82 mm). Substitute the values:\[ Z = \frac{697 - 852}{82} = \frac{-155}{82} = -1.89 \]
02

Find the cumulative probability for Z-score -1.89

Using a standard normal distribution table or a calculator, find the cumulative probability for a Z-score of -1.89. This tells us the percentage of data below 697 mm. The cumulative probability for \( Z = -1.89 \) is approximately 0.0294, or 2.94%.
03

Define the range for normal rainfall (682 mm to 1022 mm)

Normal rainfall means being within 20% of the average rainfall, which is between 682 mm and 1022 mm. We need to calculate the Z-scores for both 682 mm and 1022 mm to find the probability of rainfall within this range.
04

Calculate the Z-scores for 682 mm and 1022 mm

For 682 mm:\[ Z = \frac{682 - 852}{82} = \frac{-170}{82} = -2.07 \]For 1022 mm:\[ Z = \frac{1022 - 852}{82} = \frac{170}{82} = 2.07 \]
05

Find cumulative probabilities for Z-scores -2.07 and 2.07

Using a standard normal distribution table or a calculator, find the cumulative probabilities. For \( Z = -2.07 \), the probability is approximately 0.0192, or 1.92%. For \( Z = 2.07 \), the probability is approximately 0.9808, or 98.08%.
06

Calculate the probability for normal rainfall

Subtract the cumulative probability of the lower bound from that of the upper bound to find the percentage of years with normal rainfall: \[ P(682 < X < 1022) = 0.9808 - 0.0192 = 0.9616 \] This means that 96.16% of years will have normal rainfall.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-score
A Z-score is a way to tell how far a particular data point is from the average, in terms of the number of standard deviations. The formula to calculate a Z-score is \[ Z = \frac{X - \mu}{\sigma} \]where:
  • \(X\) is the value we are examining,
  • \(\mu\) is the mean of the distribution, and
  • \(\sigma\) is the standard deviation.
Using Z-scores, we can determine how unusual a particular measurement is within the context of a normal distribution. A Z-score of 0 indicates the data point is exactly at the mean, while positive or negative Z-scores indicate the data point is above or below the mean, respectively. For example, if India received 697 mm of rain, we can calculate its Z-score to understand how it compares to the typical rainfall. In this case, the Z-score was -1.89, indicating that 697 mm of rain is 1.89 standard deviations below the average of 852 mm. Understanding Z-scores helps us quantify unusual occurrences, like droughts, by showing how far they deviate from what is typical.
Probability
Probability is the measure of the likelihood that an event will occur. In the context of the normal distribution, probability helps us understand the chances of a particular outcome, such as receiving a specific amount of rainfall in a year. It is generally expressed as a number between 0 and 1, with 1 indicating certainty and 0 indicating impossibility.

To find the probability of receiving 697 mm or less of rain, we can use the Z-score obtained from our calculation. Since a Z-score translates an original value into how many standard deviations away it is from the mean, we can use this to look up probability in a standard normal distribution table. The probability of a Z-score of -1.89 or less (697 mm or less of rain) is approximately 0.0294. This means there is a 2.94% chance of receiving 697 mm or less rain in any given year.
Remember, probability helps us understand how likely an event or given outcome is within a certain set of parameters, like the range of rainfall over a series of years.
Cumulative Probability
Cumulative probability is the likelihood that a random variable is less than or equal to a specific value. This helps in situations where we wish to determine the probability of not just a single occurrence but any occurrence up to a certain point.

In the exercise, cumulative probability is used to determine the likelihood of having 697 mm or less rainfall in India during a monsoon season. After calculating the Z-score of -1.89 for 697 mm, we consult the standard normal distribution table to find the cumulative probability, which is 0.0294. This indicates there is a 2.94% chance of experiencing such low rainfall or less.

Cumulative probability is also useful when determining the percentage of normal rainfall years. By finding the Z-scores for both 682 mm and 1022 mm (the lower and upper bounds of what is considered normal), and then using their respective cumulative probabilities, we can calculate the probability of receiving rainfall within that range. The difference between the cumulative probabilities of the upper and lower bounds (98.08% - 1.92%) gives us the cumulative probability for normal rainfall, which turns out to be 96.16%.
  • This approach helps encapsulate the overall probability for a range of values in a distribution, making it a vital concept in statistical analysis.

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Most popular questions from this chapter

Nor mal Is Only Approximate: ACT Scores. Composite scores on the ACT for the 2019 high school graduating class had mean \(20.8\) and standard deviation \(5.8\). In all, \(1,914,817\) students in this class took the test. Of these, 227,221 had scores higher than 28 , and another 54,848 had scores exactly 28. ACT scores are always whole numbers. The exactly Normal \(N(20.8,5.8)\) distribution can include any value, not just whole numbers. What is more, there is no area exactly above 28 under the smooth Normal curve. So ACT scores can be only approximately Normal. To illustrate this fact, find a. the percentage of 2019 ACT scores greater than 28 , using the actual counts reported. b. the percentage of 2019 ACT scores greater than or equal to 28 , using the actual counts reported. c. the percentage of observations that are greater than 28 using the \(N(20.8,5.8)\) distribution. (The percentage greater than or equal to 28 is the same because there is no area exactly over 28.)

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