﻿ Standard Deviation and Normal Distribution Worksheet - Page 2 | Problems & Solutions

# Standard Deviation and Normal Distribution Worksheet - Page 2

Standard Deviation and Normal Distribution Worksheet
• Page 2
11.
The average number of hours a vehicle is in the parking lot of the company is 8. Assume the variable is normally distributed and the standard deviation is 2 hours. If 10 vehicles are selected at random, then find the probability that the mean number of hours they are kept in the parking lot will be less than 9 hours.
 a. 55.71% b. 0.9429% c. 44.29% d. 94.29%

#### Solution:

Since the variable is approximately normally distributed, the distribution of sample means will be approximately normal, with a mean of 8.

Let X be the mean number of hours that the vehicles kept in the parking lot.

The standard deviation of the sample mean is σX = σn = 210 = 0.63.

The problem is to find P(X < 9)

= P(X-μσ/n < 9 - 82/10) = P(z < 1.58)
[Convert X to standard normal units.]

= P(z < 0) + P(0 < z < 1.58) = 0.5000 + 0.4409 = 0.9429 = 94.29%.
[Use the standard normal distribution table.]

The probability that the mean number of hours that the vehicles kept in the parking lot less than 9 hours 94.29%.

12.
The average age of the employees in an organization is 34 years. Assume the variable is normally distributed and the standard deviation is 4 years. If 30 employees are selected at random, then find the probability that the mean age of the employees is greater than 32 years.
 a. 0.9938% b. 0.62% c. 99.69% d. 49.38%

#### Solution:

Since the variable is approximately normally distributed, the distribution of sample means will be approximately normal, with a mean of 34.

Let X be the sample mean age of the employees.

The standard deviation of the sample mean is σX = σn.

σX = 430 = 0.73.

The problem is to find P(X > 32)

= P(X-μσ/n > 32 - 344/30) = P(z > - 2.74)
[Convert X to standard normal units.]

= P(- 2.74 < z < 0) + P(z > 0) = 0.4969 + 0.5 = 0.9969 = 99.69%.

The probability that mean age of the employees in the sample greater than 32 years is 99.69%.

13.
The average teacher's salary in a school X is $30,000. The distribution is assumed to be normal with standard deviation equal to$5,000. If 64 teachers are selected at random, then find the probability that the mean teacher's salary is less than $28,000.  a. 99.93% b. 7% c. 49.93% d. 0.07% #### Solution: Since the variable is approximately normally distributed, the distribution of sample means will be approximately normal, with a mean of 5000. Let X be the mean teachers salary. The standard deviation of the sample mean is σX = σn = 500064 = 625. The problem is to find P(X < 28000) = P(X-μσ/n < 28000 - 300005000/64) = P(z < - 3.2) [Convert X to standard normal units.] = P(z < 0) - P(- 3.2 < z < 0) = 0.5000 - 0.4993 = 0.0007 = 0.07%. [Use the standard normal distribution table.] The probability that the mean teachers salary less than$28,000 is 0.07%.

14.
The mean time that a techician requires to perform preventive maintenance on an air-conditioning unit is 45 minutes and standard deviation is 15 minutes. The technician has to operate on 64 air conditioning units. What is the probability that the average maintenance time exceeds 40 minutes?
 a. 99.61% b. 49.61% c. 0.39% d. 0.996%

#### Solution:

Since the variable is approximately normally distributed, the distribution of sample means will be approximately normal, with a mean of 45 minutes.

Let X be the sample average maintenance time of air conditioning units.

The standard deviation of the sample mean is σX = σn = 1564 = 1.875.

The problem is to find P(X > 40)

= P(X-μσ/n > 40 - 4515/64) = P(z > - 2.66)
[Convert X to standard normal units.]

= P(- 2.66 < z < 0) + P(z > 0) = 0.4961 + 0.5000 = 0.9961 = 99.61%.
[Use the standard normal distribution table.]

The probability that sample average maintenance time of air conditioning units exceeds 40 minutes is 99.61%.

15.
The mean weight of girls in grade 12 of school X is 125 pounds, and the standard deviation is 20 pounds. If a sample of 25 girls is selected, then what is the probability that the mean of the sample will be greater than 128 pounds? Assume the variable is normally distributed.
 a. 77.34% b. 27.34% c. 0.2266% d. 22..66%

#### Solution:

Since the variable is approximately normally distributed, the distribution of sample means will be approximately normal, with a mean of 125 pounds.

Let X be the mean of sample of girls.

The standard deviation of the sample mean is σX = σn = 2025 = 4.

The problem is to find P(X > 128)

= P(X-μσ/n > 128 - 12520/25) = P(z > 0.75)
[Convert X to standard normal units.]

= P(z > 0) - P(0 < z < 0.75) = 0.5000 - 0.2374 = 0.2266 = 22.66%.
[Use the standard normal distribution table.]

The probability that the mean of the sample of girls weight will be greater than 128 pounds is 22.66%.

16.
The average protein content in different types of nuts is 20 g, and the standard deviation is 5 g. Assume the variable is normally distributed. If a sample of 16 nuts are selected, then find the probability that the mean of the sample will be greater than 22 g.
 a. 65.54% b. 34.46% c. 84.46% d. 15.54%

#### Solution:

Since the variable is approximately normally distributed, the distribution of sample means will be approximately normal, with a mean of 20g.

Let X be the mean of the sample of nuts.

The standard deviation of the sample mean is σX = σn = 2016 = 5.

The problem is to find P(X > 22)

= P(X-μσ/n > 22 - 2020/16) = P(z > 0.40)
[Convert X to standard normal units.]

= P(z > 0) - P(0 < z < 0.40) = 0.5000 - 0.1554 = 0.3446 = 34.46%.
[Use the standard normal distribution table.]

The probability that the mean protein content will be greater than 22g is 34.46%.

17.
The average price for different configurations of laptops in a shop is $1,300. Assume the standard deviation is$300. If a random sample of 9 laptops is selected, then find the probability that the mean of the sample will be less than $1,000.  a. 0.13% b. 49.87% c. 99.87% d. 13% #### Solution: Since the variable is approximately normally distributed, the distribution of sample means will be approximately normal, with a mean of$1300.

Let X be the mean price of the sample of laptops.

The standard deviation of the sample mean is σX = σn = 3009 = 100.

The problem is to find P(X < 1000)

= P(X-μσ/n < 1000 - 1300300/9) = P(z < - 3.0)
[Convert X to standard normal units.]

= P(z < 0) - P(- 3.0 < z < 0) = 0.5000 - 0.4987 = 0.0013 = 0.13%.
[Use the standard normal distribution table.]

The probability that the mean price of laptop will be less than \$1,000 is 0.13%.

18.
In a NBA playoff season the average number of points scored in a game was 200. Assume the standard deviation was 75. If a random sample of 20 games is selected, then what is the probability that the mean of the sample is lesser than 220?
 a. 17.20% b. 33.80% c. 67.20% d. 88.30%

#### Solution:

Since the variable is approximately normally distributed, the distribution of sample means will be approximately normal, with a mean of 200.

Let X be the sample mean score of the games.

The standard deviation of the sample mean is σX = σn = 7520 = 16.7.

The problem is to find P(X < 220)

= P(X-μσ/n < 220 - 20075/20) = P(z < 1.19)
[Convert X to standard normal units.]

= P(z < 0) + P(0 < z < 1.19) = 0.5000 + 0.3830 = 0.8830 = 88.30%.
[Use the standard normal distribution table.]

The probability that the mean points scored were lesser than 200 is 88.30%.

19.
In 2002 Football World Cup finals tournament a total of 64 matches were played and an average of 2.52 goals were scored per match. Assume the standard deviation to be 1.5. If a random sample of 10 games is selected, then find the probability that the mean of the sample is greater than 3 goals.
 a. 63.79% b. 13.79% c. 86.21% d. 36.21%

#### Solution:

Since the variable is approximately normally distributed, the distribution of sample means will be approximately normal, with a mean of 2.52 goals.

Let X be the sample mean of the goals scored per match.

The standard deviation of the sample means, σX = (σn)×N-nN-1, the correction factor being [N-nN-1] applied to the standard error where n is size of the sample and N is the size of the finite population.

= (σn)×N-nN-1 = (1.510)×64 - 1064 - 1 = (1.510)×5463 = 0.439
[Substitute σ = 1.5, N = 64, n = 10 and simplify.]

The problem is to find P(X > 3)

= P(X-μσX > 3 - 2.520.439) = P(z > 1.09)
[Convert X to standard normal units.]

= P(z > 0) - P(0 < z < 1.09) = 0.5000 - 0.3621 = 0.1379 =13.79%
[Use the standard normal distribution table.]

The probability that the mean goals scored were greater than 3 is 13.79%.

20.
Average score of the students in a statistics contest is 65, and the standard deviation is 15. If a random sample of 16 students who participated in the contest are selected, then find the probability that the sample mean of the scores lies between 61 and 67.
 a. 15.52% b. 20.25% c. 56.02% d. 35.77%

#### Solution:

Since the variable is approximately normally distributed, the distribution of sample means will be approximately normal with a mean of 65.

Let X be the sample mean of the scores in statistics contest.

The standard deviation of the sample mean is σX = σn = 1516 = 3.75

The problem is to find P(61 < X < 67)

= P(61 - 6515/16 < X-μσ/n < 67 - 6565/16) = P(- 1.07 < z < 0.533)
[Convert X to standard normal units.]

= P(- 1.07 < z < 0) + P(0 < z < 0.533) = 0.2025 + 0.3577 = 0.5602 = 56.02%.
[Use the standard normal distribution table.]

The probability that the sample mean of the scores in statistics contest lies between 61 and 67 is 56.02%.