﻿ Normal Distribution Word Problems | Problems & Solutions

# Normal Distribution Word Problems

Normal Distribution Word Problems
• Page 1
1.
The owner of ABC automobile Inc. credited the salaries of their staff (in the bank) for the month of May. Find the range for the shown data.
 Staff Salary ($) Manager 9000 Assistant manager 7500 Sales representative 7000 Worker1 5000 Worker2 5600 Worker3 4300  a.$6650 b. $4000 c.$4700 d. $2350 #### Solution: Range (R) = highest value - lowest value =$9000 - $4300 =$4700

2.
A survey was conducted and it was found that the number of houses in each lane of a township were 53, 109, 92, 126, 49, 223, 114, 64, 88 and 320. Find the range.
 a. 23.8 b. 135.5 c. 49 d. 271

#### Solution:

= 271
Range = highest value - lowest value = 320 - 49

So, the range of the number of houses is 271.

3.
The average price of the suburban station wagon is $20,000 with a standard deviation of$2000. The median price of the suburban station wagon is $18,000. Find the coefficient of skewness and describe the shape.  a. 0, symmetric b. - 3, negatively skewed c. + 3, positively skewed d. 3, symmetric #### Solution: Mean (X) = average price =$20,000

Median(MD) = $18,000 Standard deviation(S) =$2000

Skewness = 3(20000-18000)2000 = 3
[Skewness = 3(X-MD)S.]

The coefficient of skewness is + 3. So, the distribution is positively skewed.
[If the coefficient of skewness value is positive, then the distribution is positively skewed.]

4.
The number of personal computers sold in 10 selected show rooms of System Software Park for a particular period is 121, 326, 254, 900, 460, 1160, 85, 74, 158 and 69. Find the mean deviation.
 a. 2875.8 b. 287.58 c. 356.3

#### Solution:

Mean(X) = ΣX / n = 121+326+254+900+460+1160+85+74+158+69 / 10

= 3607 / 10 = 360.7

The deviations and absolute deviations from the mean are shown in the table.

Mean deviation = Σ(X-X)n = 2875.8 / 10 = 287.58

So, the mean deviation is 287.58.

5.
Find the sample standard deviation for the sales of Australian herbal medicines for a sample of 6 years shown. The sales (in millions dollars) are 2.1, 2.8, 3.3, 3.6, 3.8 and 4.1.
 a. 0.67 b. 0.45 c. 0.73 d. 0.53

#### Solution:

Sum of the sales, X = 2.1 + 2.8 + 3.3 + 3.6 + 3.8 + 4.1 = 19.7

Sum of the squares of the sales, X2 = 2.12 + 2.82 + 3.32 + 3.62 + 3.82 + 4.12 = 67.35

Variance of the sales, s2 = ΣX2-[(ΣX)2n]n-1

= 67.35-[(19.7)26]5

= 2.668 / 5 = 0.53

Sample standard deviation, s = s2 = 0.53 = 0.73

Therefore, the sample standard deviation for the sales is 0.73.

6.
Which of the following statement(s) is/are true?
I. The standard deviation is the square of the variance.
II. Range is not the best estimate of variability.
III. Sample variance and standard deviation are denoted by $s$2 and $s$ respectively.
IV. $\sum _{}^{}$X2 = ($\sum _{}^{}$X)2
 a. I, II and III only b. II, III and IV only c. II and III only d. I and IV only

#### Solution:

The standard deviation (σ) is the square root of the variance (σ2).
σ = σ2

Range may not be the best estimate of variability always. One extremely high or one extremely low data value can affect the range markedly.

Sample variance is denoted by s2 and the standard deviation of a sample is denoted by s.

X2 is not the same as (X)2. The notation X2 means to square the values first, then sum. (X)2 means to sum the values first, then square the sum.

7.
In a study of reaction time (in seconds) to a medicine given to 5 cancer patients, an Oncologist recorded the data 10, 20, 30, 25, 15. Find the standard deviation.
 a. 7.1 b. 50 c. 20 d. 15.81

#### Solution:

Mean, μ = ∑X / N = 10+20+30+25+155 = 20 seconds.

Compute the deviations and squares of deviations from the mean.

Variance, σ2 = Σ(X-μ)2N = 250 / 5 = 50

Standard deviation, σ = Σ(X-μ)2N = 50 = 7.1

Therefore, the standard deviation of the reaction time is 7.1.

8.
The range of a data is 2$x$ + 11. If the least score is represented by 4$x$2 - 8, then find the highest score.
 a. 4$x$2 + 4$x$ + 14 b. 4$x$2 + 2$x$ + 3 c. - 4$x$2 + 2$x$ + 19 d. 4$x$2 + 2$x$ + 19

#### Solution:

Range = highest score - lowest score

2x + 11 = highest score - (4x2 - 8), highest score = 2x + 11 + (4x2 - 8) = 4x2 + 2x + 3
[Range = 2x + 11.]

So, the highest score is represented by 4x2 + 2x + 3.

9.
What will happen to the variance and standard deviation of a data set if each numerical data is doubled?
 a. Both would be doubled. b. Variance increases by 3 times and standard deviation is doubled. c. Variance is doubled and standard deviation increases by 4 times. d. Variance increases by 4 times and standard deviation is doubled.

#### Solution:

Calculation of the variance involves squaring the deviations, the newly doubled deviations would all be squared, resulting in values that are four times as large.

The standard deviation would be doubled, since it is the square root of the variance.

So, if every number is doubled, then the variance will increase by 4 times and the standard deviation will get doubled.

10.
The amount of magazine sales (in million dollars) for a sample of 7 years are 2.6, 1.8, 2.0, 3.2, 1.6, 3.4, 2.3. Find the sample variance and sample standard deviation.
 a. 1.066 and 1.136 b. 0.475 and 0.689 c. 1.136 and 1.066 d. 2.113 and 1.45

#### Solution:

Sum of the sales, X = 2.6 + 1.8 + 2.0 + 3.2 + 1.6 + 3.4 + 2.3 = 16.9

Sum of the squares of the sales, X2 = 2.62 + 1.82 + 2.02 + 3.22 + 1.62 + 3.42 + 2.32 = 43.65

Variance of the sample, s2 = 43.65-(16.927)6
[Variance = ΣX2-(ΣX)2Nn-1.]

= 43.65-40.806 = 2.856 = 0.475

Sample standard deviation, s = variance = 0.475 = 0.689

Therefore, the sample variance for the sales is 0.475 and the sample standard deviation for the sales is 0.689.