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Basis for the Range Rule of Thumb and the Empirical Rule. In Exercises 45–48, find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. The results form the basis for the range rule of thumb and the empirical rule introduced in Section 3-2.

About______ % of the area is between z = -1 and z = 1 (or within 1 standard deviation of the mean).

Short Answer

Expert verified

The indicated area between -1 and 1 under the curve of standard normal distribution is as follows.

68.26% of the area is between z = -1 and z = 1.

Step by step solution

01

Given information

The z-scores are -1 and 1.

02

Describe the distribution

Let Z be the random variable following the standard normal distribution.

Thus,

Z~Nμ,σ2~N0,12

03

Sketch the area under the standard normal curve

Steps to draw a normal curve:

  1. Make a horizontal axis and a vertical axis.
  2. Mark the points -3.5, -3.0, -2.0 up to 3 on the horizontal axis and points 0, 0.05, 0.10 up to 0.50 on the vertical axis.
  3. Provide titles to the horizontal and vertical axes as ‘z’ and ‘P(z)’, respectively.
  4. Shade the region between -1 and 1.

The shaded area of the graph indicates the probability that the z-score is between -1 and 1.

04

Find the area of the shaded region

The area under the curve has a one-to-one correspondence with the probability.

The area between -1 and 1 is computed as

P-1<Z<1=PZ<1-PZ<-1...(1)

Referring to the standard normal table,

  • the cumulative probability of -1 is obtained from the cell intersection for rows -1.0 and the column value 0.00, which is 0.1587, and
  • the cumulative probability of 1 is obtained from the cell intersection for rows 1.0 and the column value 0.00, which is 0.8413.

Thus,

PZ<1=0.8413PZ<-1=0.1587

Substituting in equation (1),

P-1<Z<1=PZ<1-PZ<-1=0.8413-0.1587=0.6826

Thus, the area between -1 and 1 is 0.6826.

05

Express the result in percentage

The area can be expressed as

0.6826×100=68.26%

Thus,about 68.26 % of the area is between z = -1 and z = 1.

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Most popular questions from this chapter

In Exercises 13–20, use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.)

Mean

St.Dev.

Distribution

Males

23.5 in

1.1 in

Normal

Females

22.7 in

1.0 in

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Find the probability that a female has a back-to-knee length greater than 24.0 in.

Durations of PregnanciesThe lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.

a. In a letter to “Dear Abby,” a wife claimed to have given birth 308 days after a brief visit fromher husband, who was working in another country. Find the probability of a pregnancy lasting308 days or longer. What does the result suggest?

b. If we stipulate that a baby is prematureif the duration of pregnancy is in the lowest 3%,find the duration that separates premature babies from those who are not premature. Premature babies often require special care, and this result could be helpful to hospital administrators in planning for that care.

In Exercises 7–10, use the same population of {4, 5, 9} that was used in Examples 2 and 5. As in Examples 2 and 5, assume that samples of size n = 2 are randomly selected with replacement.

Sampling Distribution of the Sample Standard Deviation For the following, round results to three decimal places.

a. Find the value of the population standard deviation

b. Table 6-2 describes the sampling distribution of the sample mean. Construct a similar table representing the sampling distribution of the sample standard deviation s. Then combine values of s that are the same, as in Table 6-3 (Hint: See Example 2 on page 258 for Tables 6-2 and 6-3, which describe the sampling distribution of the sample mean.)

c. Find the mean of the sampling distribution of the sample standard deviation. d. Based on the preceding results, is the sample standard deviation an unbiased estimator of the population standard deviation? Why or why not?

Standard Normal DistributionIn Exercises 17–36, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers

to four decimal places.

Greater than 0.25

Significance For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that are

a. significantly high (or at least 2 standard deviations above the mean).

b. significantly low (or at least 2 standard deviations below the mean).

c. not significant (or less than 2 standard deviations away from the mean).

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