# Standard Deviation and Normal Distribution Worksheet

Standard Deviation and Normal Distribution Worksheet
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1.
Assume that the weights of 3,000 students at a university are normally distributed with mean 150 pounds and a standard deviation of 7 pounds. If 100 samples of 30 students each are obtained, then what would be the expected mean and standard deviation?
 a. 150 lbs, $\frac{7}{10}$ lbs b. 150 lbs, $\frac{7}{30}$ lbs c. 150 lbs, $\frac{7}{3000}$ lbs d. 150 lbs, 1.278 lbs

#### Solution:

Since the weights in the population are normally distributed with mean 150 lbs and standard deviation 7 lbs, the distribution of sample means will be approximately normal with a mean of 150 lbs.

The standard deviation of the sample means is σx = σn where σ is the standard deviation of the population and n = size of the sample drawn.

σx == 730 lbs = 1.278 lbs.
[Substitute n = 30 and σ = 7.]

2.
Assume that the weights of 3,000 students at a university are normally distributed with mean 150 pounds and a standard deviation of 7 pounds. If 100 samples of 30 students each are obtained, then what would be the expected mean and standard deviation of the resulting sampling distribution of means if sampling were done without replacement?
 a. 150 lbs, $\frac{7}{3000}$ lbs b. 150 lbs, $\frac{7}{10}$ lbs c. 150 lbs, $\frac{7}{30}$ lbs d. 150 lbs, 1.272 lbs

#### Solution:

Here the distribution of sample means will be approximately normal with a mean of 150 lbs.

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

The standard deviation of the sample means = (σn)×N-nN-1 = (730)×3000-303000-1 = 1.278 × 0.99515 = 1.272 lbs

3.
The average annual salary in a district is $30,000. Assuming that the salaries were normally distributed for a certain group of wage earners with a standard deviation of$5,000, what is the probability that for a randomly selected sample of 70 earners the mean salary falls below $28,000?  a. 0.01% b. 0.04% c. 4% d. 0.1% #### Solution: The population mean =$30,000 and standard deviation of the population = $5,000. The sample of 70 earners drawn from the population will follow a normal distribution with the mean$30,000 and standard deviation, σx = 500070 = 597.614.
[Use σx = σn.]

The salary of $28,000 in standard units z = ($28000-$30000)597.614 = - 2000 / 597.614 = - 3.347 The area of the normal curve between z = - 3.347 and z = 0 is 0.4996. [Use the standard normal distribution table.] So, the area of the normal curve for z < - 3.347 is 0.5 - 0.4996 = 0.0004 That is the probability that the sample mean falls below$28,000 is 0.04%.

4.
The average age of media reporters worldwide is 35 years with a standard deviation of 6 years. If a media channel employs 50 media reporters, then what is the probability that their mean age lies between 34 and 37 years? Assume that the variable is normally distributed.
 a. 87.15% b. 75% c. 80% d. 90 %

#### Solution:

The sample of 50 media reporters drawn from the population will follow a normal distribution with the mean age of 35 years and standard deviation, σx = 650 = 0.84853
[Use σx = σn.]

Here the mean x of the sample in standard normal units is given by, z = (x-35)0.84853.

The age 34 years in standard units, z = (34-35)0.84853 = - 1.1785

The age 37 years in standard units, z = (37-35)0.84853 = 2.357

Probability of samples with mean age between 34 and 37 = the area under normal curve between z = - 1.1785 and z = 2.357 is 0.3807 + 0.4908 = 0.8715.
[Use the standard normal distribution table.]

The probability that the mean age of sample of 50 media reporters is 87.15%.

5.
The average age of media reporters worldwide is 35 years with a standard deviation of 6 years. If a media channel employs 50 media reporters, then what is the probability that their mean age is above 38 years? Assume that the variable is normally distributed.
 a. 0.1% b. 1% c. 0.02% d. 2%

#### Solution:

The sample of 50 media reporters drawn from the population will follow a normal distribution with the mean age of 35 years and standard deviation, σx = 650 = 0.84853.
[Use σx = σn.]

Age 38 in standard units, z = (38-35)0.84853 = 3.5355.

The area of the normal curve between z = 0 and z = 3.5355 is 0.4998.

The area of the normal curve for z > 3.5355 is 0.5 - 0.4998 = 0.0002.

Hence the probability that the mean age of sample to be above 38 years is 0.0002 or 0.02%.

6.
A property value assessor has appraised that the average value of 600 houses in an area was $250,000 and the standard deviation was$80,000. If 50 houses are marked for sale, then find the probability that the mean of the values of these houses is greater than $275,000.  a. 1.6% b. 1.06% c. 6.01% d. 6.1% #### Solution: The standard deviation of the sample means is given by (σn)×N-nN-1, the correction factor being [N-nN-1] applied to the standard error where n is the size of the sample and N is the size of the finite population. The standard deviation of the sample means = (σn)×N-nN-1 = (8000050)×600-50600-1 = 10841.09. The mean x of the sample in standard normal units is, z = (x-250000)10841.09. The price of$275,000 in standard units, z = (275000-250000)10841.09 = 2.306.

Probability of samples with mean price $275,000 = the area under normal curve z = 0 and z = 2.306 is 0.4894. [Use the standard normal distribution table.] So, the area under normal curve for z > 2.306 is 0.5 - 0.4894 = 0.0106. That is, the proprobability that the mean of the values of the houses is greater than$275,000 is 1.06%.

7.
In a study of mortality rates of 500 tribals in a region, the mean life expectancy was found to be 73 years, with a standard deviation of 6 years. If a sample of 50 tribals is selected at random, then find the probability that their life expectancy is less than 72 years.
 a. 3.93% b. 39.3% c. 10.73% d. 7.01%

#### Solution:

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

The standard deviation of the sample means = (σn)×N-nN-1 = (650)×500-50500-1 = 0.8058.

The mean x of the sample in standard normal units is, z = (x-73)0.8058.

So, 72 years in standard units, z = (72-73)0.8058 = - 1.241.

The area of the curve between z = - 1.241 and z = 0 is 0.3927.
[Use the standard normal distribution table.]

The area of the normal curve for z < - 1.241 is 0.5 - 0.3927 = 0.1073

Probability that the life expectancy is less than 72 years is 0.1073 or 10.73%.

8.
Assume that the weights of 3,000 students at a university are normally distributed with mean 150 pounds and a standard deviation of 2.72 pounds. How many samples would you expect to find the mean between 147 and 155 pounds?
 a. 83 b. 82 c. 84 d. 81

#### Solution:

The mean x of a sample in standard normal units is, z = (x-μ)σ = (x-150)2.72.

Convert 147 and 155 into standard normal units.

For x = 147, z1 = (147-150)2.72 = - 1.103
For x = 155, z2 = (155-150)2.72 = 1.84

Proportion of samples with means between 147 and 155 = the area under normal curve between z1 = - 1.103 and z2 = 1.84

= (the area of the normal curve between z = - 1.103 and z = 0) + (the area of the normal curve between z = 0 and z = 1.84) = 0.3708 + 0.4671 = 0.8379.
[Use the standard normal distribution table.]

Expected number of samples = 100 × 0.8379 = 83.79 (or) 84.

9.
Assume that the weights of 3,000 students at a university are normally distributed with mean 150 pounds and a standard deviation of 2.72 pounds. How many samples would you expect to find the mean below 146 pounds?
 a. 7 b. 6 c. 5 d. 8

#### Solution:

The variable x = 146 in standard units, z = (146-150)2.72 = - 1.4706.
[Use z = (x-μ)σ.]

The area of the normal curve between z = - 1.4706 and z = 0 is 0.43.
[Use the standard normal distribution table.]

The area of the normal curve z < - 1.4706 is 0.5 - 0.43 = 0.07.

So, the expected number of samples for the mean below 146 pounds is 100 × 0.07 = 7.

10.
According to a school report, the students in that school spend on an average 15 hours in front of computers. Assume the variable is normally distributed and the standard deviation is 2 hours. If 30 students are selected at random, then find the probability that the mean of the number of hours they spend in front of computers will be greater than 16 hours.
 a. 0.31% b. 31% c. 49.9% d. 0.0031%

#### Solution:

Since the variable is approximately normally distributed, the distribution of sample means will be approximately normal, with a mean of 15.

Let X be the mean of the number of hours spend in front of computers.

The standard deviation of the sample mean is σX = σn = 230 = 0.365.

The problem is to find P(X > 16).

= P(X-μσ/n > 16 - 152/30) = P(z > 2.74)
[Convert X to standard normal units.]

= P(z > 0) - P(0 < z < 2.74) = 0.5000 - 0.4969 = 0.0031 = 0.31% .
[Use the standard normal distribution table.]

The probability of obtaining a sample mean larger than 16 hours is 0.31%.