Table A practice

(a) z<2.85

(b) z>2.85

(c) z>-1.66

(d) -1.66<z<2.85

Short Answer

Expert verified

a). The proportion is read as0.9978.

b). The area to the right of z=2.85 is .0022

c). The probability of P(z>-1.66)is.9515.

d). Area (proportion) B is .9493, which is therole="math" localid="1652870190253" P(-1.66<z<2.85).

Step by step solution

01

Part (a) Step 1: Given Information

Given data:

z<2.85

02

Part (a) Step 2: Explanation

To determine each proportion corresponding to a given z value, we will use the Standard Normal table. The notation P()indicates that we want to compute a probability or a proportion.

P(z<2.85). Here's a sketch of the intended proportion, with the shaded area of interest:

We find the z-value 2.8on the left side of the Standard Normal table, then proceed to the right in that row until we reach the column headed by 05. 0.9978.

03

Part (b) Step 1: Given Information

Given data:

z>-1.66.

04

Part (b) Step 2: Explanation

This is the "opposite" of the percentage we discovered in part one (A). We shade the typical normal curve on the other side as follows:

05

Part (c) Step 1: Given Information

Given data:

z>-1.66

06

Part (c) Step 2: Explanation

P(z>-1.66)Here is the sketch of the desired proportion:

From the Standard Normal table, we look up the z-value -1.6on the left side of the table, and then move to the right in that row until we are under the column headed by .06. The proportion is read as .0485. But this is the proportion to the left of -1.66. To calculate the proportion to theright of -1.66, we simply subtract from 1 as shown here:

Area to right 1-0.485

=.9515

07

Part (d) Step 1: Given Information

Given data:

-1.66<z<2.85

08

Part (d) Step 2: Explanation

P(-1.66<z<2.85).This is a "between" problem that will take several steps to solve. First, we'll make the sketch:

We now look at the sketch again, and we identify three different pieces of it: the unshaded piece to the left of -1.66, the shaded piece between -1.66and 2.85, and the unshaded piece to the right of 2.85.

09

Part (d) Step 3: Explanation

We designate them as areas A. B, and C, respectively:

In part (C) we determined that area A was .0485, and in part (B) we found that area C was .0022.

We know that the total area, or proportion, under the curve is 1 . Thus, we know that A+B+C=1. We know A and C already. So it is a matter of subtraction to determine the value of area B, which is the area (proportion) of interest.

A+B+C=1

.0485+B+.0022=1

B+.0507=1

B=.9493

Area (proportion) B is .9493, which is the P(-1.66<z<2.85).

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Most popular questions from this chapter

T2.2. For the Normal distribution shown, the standard deviation is closest to

(a) 0

(b) 1

(c) 2

(d) 4

(e) 5

R2.9 low-birth-weight babies Researchers in Norway analyzed data on the birth weights of 400,000 newborns over a 6-year period. The distribution of birth weights is Normal with a mean of 3668 grams and a standard deviation of 511 grams. 17 Babies that weigh less than 2500 grams at birth are classified as "low birth weight."

(a) What percent of babies will be identified as having low birth weight? Show your work.

(b) Find the quartiles of the birth weight distribution. Show your work.

T2.5. The average yearly snowfall in Chillyville is Normally distributed with a mean of 55 inches. If the snowfall in Chillyville exceeds 60inches in 15% of the years, what is the standard deviation?
(a) 4.83 inches
(d) 8.93 inches
(b) 5.18 inches
(c) The standard deviation
(c) 6.04 inches cannot be computed from the given information.

Use Table A to find the proportion of observations from the standard Normal distribution that satisfies each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question. Use your calculator or the Normal Curve applet to check your answers.

More Table A practice

(a) zis between −1.33and 1.65

(b) zis between 0.50and1.79

Normal is only approximate: ACT scores Scores on the ACT test for the 2007 high school graduating class had mean 21.2and standard deviation 5.0. In all, 1,300,599students in this class took the test. Of these, 149,164had scores higher than 27and another 50,310had scores exactly 27. ACT scores are always whole numbers. The exactly Normal N(21.2,5.0)distribution can include any value, not just whole numbers. What’s more, there is no area exactly above 27under the smooth Normal curve. So ACT scores can be only approximately Normal. To illustrate this fact, find

(a) the percent of 2007ACT scores greater than 27.

(b) the percent of 2007ACT scores greater than or equal to 27.

(c) the percent of observations from theN(21.2,5.0) distribution that are greater than 27. (The percent greater than or equal to 27 is the same, because there is no area exactly over 27.)

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