Standard Deviation and Normal Distribution Worksheet

**Page 1**

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 |

The standard deviation of the sample means is σ

σ

[Substitute

Correct answer : (4)

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 |

The standard deviation of the sample means is given by

The standard deviation of the sample means =

Correct answer : (4)

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% |

The sample of 70 earners drawn from the population will follow a normal distribution with the mean $30,000 and standard deviation, σ

[Use σ

The salary of $28,000 in standard units

The area of the normal curve between

[Use the standard normal distribution table.]

So, the area of the normal curve for

That is the probability that the sample mean falls below $28,000 is 0.04%.

Correct answer : (2)

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 % |

[Use σ

Here the mean

The age 34 years in standard units,

The age 37 years in standard units,

Probability of samples with mean age between 34 and 37 = the area under normal curve between

[Use the standard normal distribution table.]

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

Correct answer : (1)

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% |

[Use σ

Age 38 in standard units,

The area of the normal curve between

The area of the normal curve for

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

Correct answer : (3)

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% |

The standard deviation of the sample means =

The mean

The price of $275,000 in standard units,

Probability of samples with mean price $275,000 = the area under normal curve

[Use the standard normal distribution table.]

So, the area under normal curve for

That is, the proprobability that the mean of the values of the houses is greater than $275,000 is 1.06%.

Correct answer : (2)

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% |

The standard deviation of the sample means =

The mean

So, 72 years in standard units,

The area of the curve between

[Use the standard normal distribution table.]

The area of the normal curve for

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

Correct answer : (3)

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 |

Convert 147 and 155 into standard normal units.

For

For

Proportion of samples with means between 147 and 155 = the area under normal curve between

= (the area of the normal curve between

[Use the standard normal distribution table.]

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

Correct answer : (3)

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 |

[Use

The area of the normal curve between

[Use the standard normal distribution table.]

The area of the normal curve

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

Correct answer : (1)

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% |

Let

The standard deviation of the sample mean is

The problem is to find P(

= P(

[Convert

= P(

[Use the standard normal distribution table.]

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

Correct answer : (1)