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New York City Weather New York City's mean minimum daily temperature in February is \(27^{\circ} \mathrm{F}\) (http://www.ny.com). Suppose the standard deviation of the minimum temperature is \(6^{\circ} \mathrm{F}\) and the distribution of minimum temperatures in February is approximately Normal. What percentage of days in February has minimum temperatures below freezing \(\left(32^{\circ} \mathrm{F}\right) ?\)

Short Answer

Expert verified
There is approximately 79.77% chance that the minimum temperature on a given February day in New York City will be below freezing.

Step by step solution

01

Understand the Problem

The problem is asking for the percentage of days in February with minimum temperatures below freezing \(32^{\circ} \mathrm{F}\). The information given is that the mean minimum temperature in February is \(27^{\circ} \mathrm{F}\) and the standard deviation is \(6^{\circ} \mathrm{F}\). The temperature distribution is assumed to be normal.
02

Calculate the Z-score

Next, calculate the Z-score for \(32^{\circ} \mathrm{F}\), the freezing point. The Z-score is calculated by subtracting the mean from the value, then dividing by the standard deviation. Therefore, \(Z = \frac{32-27}{6} = 0.83\)
03

Find the Percentage

Finally, look up the Z-score in a standard normal distribution table to find the area to the left (since we're interested in temperatures below \(32^{\circ} \mathrm{F}\)). If the Z-score is 0.83, the area (or probability) to the left is 0.79767, indicating that there is approximately 79.77% chance that the minimum temperature on a given February day in New York City will be below freezing.

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Key Concepts

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

Z-score Calculation
Understanding the Z-score is crucial when working with normal distributions. In essence, the Z-score represents the number of standard deviations a data point is from the mean. When we measure weather temperatures, like the chilly days of New York City's February we are analyzing, calculating the Z-score of a specific temperature can tell us how common or rare that temperature is compared to what is typical (the mean).

Let's see how this is done. To calculate the Z-score, you subtract the mean value from the data point you're interested in. For our example, we are comparing the freezing point of water, which is \(32^{\text{o}} \text{F}\), to the mean temperature. Next, you divide this difference by the standard deviation. The formula for calculating a Z-score looks like this: \[ Z = \frac{\text{data point} - \text{mean}}{\text{standard deviation}} \].

Applying this formula to our problem where the mean temperature is \(27^{\text{o}} \text{F}\) with a standard deviation of \(6^{\text{o}} \text{F}\), the Z-score is calculated as \( Z = \frac{32 - 27}{6} = 0.83 \). This Z-score helps us to understand where the temperature of \(32^{\text{o}} \text{F}\) falls in relation to the 'average' February day in New York City.
Standard Normal Distribution Table
To grasp the concept of the standard normal distribution table, it is important to know why it is used. Once you've computed a Z-score, the next step is to figure out what this number actually means in terms of probability. This is where the standard normal distribution table comes into play.

Each Z-score corresponds to a percentage which represents the area under the normal curve to the left of that Z-score. This area signifies the probability that a value in our normal distribution is less than our data point. For our problem with the freezing temperatures, after calculating a Z-score of 0.83, we consult the table to find out the percentage of days that have temperatures below \(32^{\text{o}} \text{F}\).

Looking up a Z-score of 0.83 in the standard normal distribution table, we find that approximately 79.77% of the area under the curve falls to the left of this Z-score. This means that there's roughly a 79.77% chance that a randomly chosen February day in New York City will have a minimum temperature below freezing. Such tables are extensively used in various fields such as meteorology, economics, and health sciences to interpret Z-scores into more meaningful probabilities.
Mean and Standard Deviation
The mean and standard deviation are fundamental statistical concepts that play a significant role in understanding data, like the weather patterns we're examining. The mean is simply the average of all the data points. It's the balance point of the dataset. In the context of New York City's February weather, the mean minimum daily temperature is given as \(27^{\text{o}} \text{F}\). It represents the 'center' of the temperature data.

The standard deviation, on the other hand, measures how spread out the numbers in your data are. A small standard deviation means the data points tend to be close to the mean, whereas a large standard deviation means they are spread out over a wider range of values. Here, the standard deviation for the minimum temperature is \(6^{\text{o}} \text{F}\), suggesting that daily minimum temperatures in February typically vary by about \(6^{\text{o}} \text{F}\) from the average. Together, these two parameters shape the normal distribution bell curve, and ultimately, they are critical in determining probabilities like those we are calculating for temperatures below freezing.

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Most popular questions from this chapter

The distribution of the math portion of SAT scores has a mean of 500 and a standard deviation of 100 , and the scores are approximately Normally distributed. a. What is the probability that one randomly selected person will have an SAT score of 550 or more? b. What is the probability that four randomly selected people will all have SAT scores of 550 or more? c. For 800 randomly selected people, what is the probability that 250 or more will have scores of 550 or more? d. For 800 randomly selected people, on average how many should have scores of 550 or more? Round to the nearest whole number. e. Find the standard deviation for part d. Round to the nearest whole number. f. Report the range of people out of 800 who should have scores of 550 or more from two standard deviations below the mean to two standard deviations above the mean. Use your rounded answers to part \(\mathrm{d}\) and \(\mathrm{e}\). g. If 400 out of 800 randomly selected people had scores of 550 or more, would you be surprised? Explain.

Use a table or technology to answer each question. Include an appropriately labeled sketch of the Normal curve for each part. Shade the appropriate region. a. Find the probability that a z-score will be \(1.76\) or less. b. Find the probability that a z-score will be \(1.76\) or more. c. Find the probability that a \(z\) -score will be between \(-1.3\) and \(-1.03\).

Use a table or technology to answer each question. Include an appropriately labeled sketch of the Normal curve for each part. Shade the appropriate region. a. Find the area to the left of a \(z\) -score of \(-0.50\). h. Find the area to the right of a \(z\) -score of \(-0.50\).

ACT scores are approximately Normally distributed with a mean of 21 and a standard deviation of 5, as shown in the figure. (ACT scores are test scores that some colleges use for determining admission.) What is the probability that a randomly selected person scores 24 or more?

The average birth weight of elephants is 230 pounds. Assume that the distribution of birth weights is Normal with a standard deviation of 50 pounds. Find the birth weight of elephants at the 95 th percentile.

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