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Chapter 6: Problem 6

Find the area under the standard normal curve to the left of \(z=-1.37.\)

Short Answer

Expert verified
The area under the standard normal curve to the left of \(z=-1.37\) can be found using a standard normal distribution table by looking up the value corresponding to \(z=1.37\), due to symmetry. The exact value will be in the z-table, representing the cumulative probability.

Step by step solution

01

Notation

Firstly, It's important to note that for the standard normal curve, the mean is at \(0\) and the standard deviation is \(1\). The variable \(Z\) represents a standard score or 'z-score.'. The value \(z = -1.37\) positions 'z' a little more than one standard deviation to the left of the mean '0'.
02

Read from the Z-table

The Z-table (Standard Normal Distribution Table) gives you the area to the left of a positive 'Z' value. But because of symmetry, the area to the left of \(-1.37\) is the same as the area to the right of \(1.37\). Look up \(1.37\) on the standard normal table.
03

Interpretation

The value found in the table will be the desired cumulative probability. This is the area under the curve to the left of \(z=-1.37\).

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Key Concepts

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

Z-Score
Understanding z-score is critical when studying statistics, as it is a measurement indicating how many standard deviations an element is from the mean. A z-score is a numerical representation that describes a value's relationship to the mean of a group of values, standardized so that it's applicable to different data sets.

In the exercise, the given z-score was \(z = -1.37\). This implies that the score is 1.37 standard deviations below the mean of the distribution (since it鈥檚 negative). The z-score enables comparison across different normal distributions, which is particularly powerful for statistical analysis.
Standard Deviation
The term standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean (also called the expected value) of the set, while a high standard deviation indicates that the values are spread out over a wider range.

As applied to the standard normal distribution, the standard deviation is always 1. This is a crucial concept because it allows us to use the standard normal curve and tables to determine probabilities and areas under the curve for any normal distribution, once values have been standardized, thus transforming them into z-scores.
Normal Distribution
The normal distribution, often termed the bell curve, is a probability distribution that is symmetric about the mean. It exhibits the property that a majority of the data points are relatively similar, manifesting in most outcomes clustering around the average value.

In the exercise, we're dealing with the standard normal distribution, a special case of the normal distribution where the mean is 0 and the standard deviation is 1. It's important to note the curve's distinct characteristic: it is symmetrical, meaning the left half is a mirror image of the right half.
Cumulative Probability
The concept of cumulative probability refers to the likelihood that a random variable takes on a value less than or equal to a specific value. In the context of the standard normal curve, it represents the area under the curve to the left of a certain z-score.

In the given exercise, the cumulative probability was sought for \(z = -1.37\). Using the Z-table, which lists the cumulative probabilities from the left up to a z-score, we can find the likelihood of obtaining a value to the left of our specific z-score, equivalent to the stated area under the curve.

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Most popular questions from this chapter

Given that \(x\) is a normally distributed random variable with a mean of 60 and a standard deviation of 10 find the following probabilities. a. \(\quad P(x>60)\) b. \(\quad P(60< x<72)\) c. \(\quad P(57< x<83)\) d. \(\quad P(65< x<82)\) e. \(\quad P(38< x<78)\) f. \(\quad P(x<38)\)

The data below are the net weights (in grams) for a sample of 30 bags of \(\mathrm{M} \& \mathrm{M}\) 's. The advertised net weight is 47.9 grams per bag. $$\begin{array}{llllll}\hline 46.22 & 46.72 & 46.94 & 47.61 & 47.67 & 47.70 \\\47.98 & 48.28 & 48.33 & 48.45 & 48.49 & 48.72 \\\48.74 & 48.95 & 48.98 & 49.16 & 49.40 & 49.69 \\\49.79 & 49.80 & 49.80 & 50.01 & 50.23 & 50.40 \\\50.43 & 50.97 & 51.53 & 51.68 & 51.71 & 52.06 \\\\\hline\end{array}$$ The FDA requires that (nearly) every bag contain the advertised weight; otherwise, violations (less than 47.9 grams per bag) will bring about mandated fines. (M\&M's are manufactured and distributed by Mars Inc. a. What percentage of the bags in the sample are in violation? b. If the weight of all filled bags is normally distributed with a mean weight of 47.9 g, what percentage of the bags will be in violation? c. Assuming the bag weights are normally distributed with a standard deviation of \(1.5 \mathrm{g},\) what mean value would leave \(5 \%\) of the weights below \(47.9 \mathrm{g} ?\) d. Assuming the bag weights are normally distributed with a standard deviation of \(1.0 \mathrm{g},\) what mean value would leave \(5 \%\) of the weights below \(47.9 \mathrm{g} ?\) e. Assuming the bag weights are normally distributed with a standard deviation of \(1.5 \mathrm{g},\) what mean value would leave \(1 \%\) of the weights below \(47.9 \mathrm{g} ?\) f. Why is it important for Mars to keep the percentage of violations low? g. It is important for Mars to keep the standard deviation as small as possible so that in turn the mean can be as small as possible to maintain net weight. Explain the relationship between the standard deviation and the mean. Explain why this is important to Mars.

The 70 -year long-term record for weather shows that for New York State, the annual precipitation has a mean of 39.67 inches and a standard deviation of 4.38 inches [Department of Commerce; State, Regional and National Monthly Precipitation Report]. If the annual precipitation amount has a normal distribution, what is the probability that next year the total precipitation will be: a. more than 50.0 inches? b. between 42.0 and 48.0 inches? c. between 30.0 and 37.5 inches? d. more than 35.0 inches? e. less than 45.0 inches? f. less than 32.0 inches?

Findings from a survey of American adults conducted by Yankelovich Partners for the International Bottled Water Association indicate that Americans on the average drink 6.18-ounce servings of water a day [http://www.pangaeawater.com/]. Assuming that the number of 8-ounce servings of water is approximately normally distributed with a standard deviation of 1.4 servings, what proportion of Americans drink a. more than the recommended 8 servings? b. less than half the recommended 8 servings?

a. Find the area under the standard normal curve to the left of \(z=0, P(z<0).\) b. Find the area under the standard normal curve to the right of \(z=0, P(z>0).\)
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