Return to the definitions in Class Handout #2 to see some properties of a normal distribution, which is one type of bell-shaped (mound shaped) distribution that is very important in statistics:

+ – – 2 + 2 – 3 + 3

normal distribution a symmetric and bell-shaped distribution which can be described entirely in terms of the number of standard deviations () that values are away from the mean ()

68.3%

95.4%

99.7%

The normal distribution is very important in statistics when we study sampling distributions.

parent population the population from which a sample is to be selected

x is used to represent the mean of the population consisting of the possible values of x from a random sample size n, that is, the mean of the sampling distribution of x with random samples of size n.

x is used to represent the standard deviation of the population consisting of the possible values of x from a random sample size n, that is, the standard deviation of the sampling distribution of x with random samples of size n.

x is called the standard error of the mean.

sampling distribution of x a probability distribution for all values of y that are possible with (random) samples of size n

Class Exercise #4 illustrates why the normal distribution is so important when we study sampling distributions:

4.

(a)

(b)

Consider an ordinary, six-sided die, that is, one with six sides where one side has a single spot, a second side has two spots, a third side has three spots, a fourth side has four spots, a fifth side has five spots, and the sixth side has six spots. We assume the die is fair, that is, each side is equally likely to face up with each roll.

Suppose the die is rolled 100 times, and the number of spots facing up is recorded for each roll. Sketch a histogram which would most resemble a histogram for this data.

Suppose 10 such dice are all rolled simultaneously and the mean number of spots for the 10 dice is recorded for each of the simultaneous rolls. We could consider each of the 10 simultaneous rolls as a random sample of size n = 10. Identify the parent population and the sampling distribution.

Uniform Distribution

The parent population consists of the numbers 1, 2, 3, 4, 5, 6. The sampling distribution is the probability distribution for all values of the mean for 10 simultaneous rolls that are possible.

(c) Suppose 100 simultaneous rolls of the 10 dice in part (b) are made and the mean number of spots is recorded for each of the simultaneous rolls. Sketch a histogram which would most resemble a histogram for this data.

Bell-Shaped Distribution

It is very unlikely that the mean for 10 simultaneous rolls will be equal to values near 1 or values near 6. It is very likely that the mean for 10 simultaneous rolls will be equal to values near 3 or 4.

Part (c) illustrates why the normal distribution is so important when we study sampling distributions. The Central Limit Theorem, found with the definitions in Class Handout #2, is a more precise statement of the properties of sampling distributions:

When selecting a simple random sample of size n from a parent population which can be treated as an infinite population with mean and standard deviation , then the following are true:

(1)

(2)

(3)

The mean of the sampling distribution of x is

The standard deviation of the sampling distribution of x is

For a sufficiently large sample size n, the sampling distribution of y will be

(or at least can be accurately approximated by) a normal distribution;

what sample size n will be sufficiently large depends on how close the

distribution of the parent population is to a normal distribution.

x = .

x = —— .

n

Central Limit Theorem a statement of the following fact (requiring advanced mathematics to prove):

We now return to part (d) of Class Exercise #4.

(c)

(d)

Suppose 100 simultaneous rolls of the 10 dice in part (b) are made and the mean number of spots is recorded for each of the simultaneous rolls. Sketch a histogram which would most resemble a histogram for this data.

Bell-Shaped Distribution

What does the Central Limit Theorem say about how the probability distribution for a single roll (in part (a)) and the probability distribution for 10 simultaneous rolls (in part (c)) compare with each other?

The mean is the same (3.5) for the two distributions.

It is very unlikely that the mean for 10 simultaneous rolls will be equal to values near 1 or values near 6. It is very likely that the mean for 10 simultaneous rolls will be equal to values near 3 or 4.

The standard deviation for the distribution of the mean of 10 simultaneous rolls is smaller than the standard deviation for the distribution of a single roll.

The distribution of the mean of 10 simultaneous rolls is more like a normal distribution than the distribution of a single roll.

+ – – 2 + 2 – 3 + 3

z-score (standard score) the number of standard deviations a value x in a population is away from the mean ()

x – z = ——

normal distribution a symmetric and bell-shaped distribution which can be described entirely in terms of the number of standard deviations () that values are away from the mean ()

68.3%

95.4%

99.7%

Table 2 of the Statistical Tables (available from the appropriate link on the web page syllabus) displays probabilities corresponding to various z-scores (for what is often called the standard normal distribution).

5.

(a)

On July 31, 1998, the information below about mean and standard deviation of heights was obtained via the Anthropometric Desk Reference web page at

http://www.odc.com/anthro/deskref/desktoc.htmlWe assume that heights are normally distributed for each age group and sex.

Population Mean Height (cm.) Standard Deviation of Height (cm.)Four-Year-Old Girls 101.6 4.0Four-Year-Old Boys 102.9 4.3Nine-Year-Old Girls 132.2 6.5Nine-Year-Old Boys 132.2 5.7

Draw a sketch to display the probability distribution curve for the heights of nine-year old girls.

132.2

68.3%

99.7%

95.4%

138.7125.7 145.2119.2 151.7112.7

Draw a sketch to display the probability that one randomly selected four-year-old girl has a height greater than 102.2 cm. Then, use Table 2 of the Statistical Tables to find this probability.

= 101.6 = 4.0

101.6

102.2

z = (102.2 – 101.6) / 4.0 = +0.15

(z = +0.15)

0.4404 or 44.04%

(b)

Draw a sketch to display the probability that one randomly selected four-year-old girl has a height less than 102.2 cm. Then, use Table 2 of the Statistical Tables to find this probability.

(c)

101.6

= 101.6 = 4.0

z = (102.2 – 101.6) / 4.0 = +0.15

1 – 0.4404 = 0.5596 or 55.96%

102.2 (z = +0.15)

5.-continued

Draw a sketch to display the probability that one randomly selected four-year-old boy has a height less than 101.6 cm. Then, use Table 2 of the Statistical Tables to find this probability.

(d)

102.9

= 102.9 = 4.3

101.6

z = (101.6 – 102.9) / 4.3 = –0.30

(z = –0.30)

0.3821 or 38.21%

5.-continuedDraw a sketch to display the probability that one randomly selected four-year-old boy has a height greater than 100 cm. Then, use Table 2 of the Statistical Tables to find this probability.

102.9

= 102.9 = 4.3

100

z = (100 – 102.9) / 4.3 = –0.67

(z=–0.67)

1 – 0.2514 = 0.7486 or 74.86%

(e)

Draw a sketch to display the probability that one randomly selected nine-year-old girl has a height between 126 and 140 cm. Then, use Table 2 of the Statistical Tables to find this probability.

(f)

= 132.2 = 6.5

132.2126

z = (126 – 132.2) / 6.5 = –0.95 z = (140 – 132.2) / 6.5 = +1.20

140

0.1711 0.1151

1 – (0.1711 + 0.1151) = 0.7138 or 71.38%

(z=–0.95) (z=+1.20)

Draw a sketch to display the probability that one randomly selected nine-year-old girl has a height between 135 and 145 cm. Then, use Table 2 of the Statistical Tables to find this probability.

5.-continued (g)

132.2

= 132.2 = 6.5

0.3336 – 0.0244 = 0.3092 or 30.92%

135

0.3336

0.0244

145

(z=+0.43)

(z=+1.97)

Draw a sketch to display the probability that one randomly selected nine-year-old girl has a height between 125.5 and 129.5 cm. Then, use Table 2 of the Statistical Tables to find this probability.

(h)

= 132.2 = 6.5

132.2

0.3372 – 0.1515 = 0.1857 or 18.57%

125.5

0.3372

0.1515

129.5(z = –1.03) (z = –

0.42)

Draw a sketch to display the probability that one randomly selected nine-year-old boy has a height less than the mean height for four-year-old girls. Then, use Table 2 of the Statistical Tables to find this probability.

132.2

= 132.2 = 5.7

101.6

practically zero (0)

(z = –5.37)

5.-continued (i)

Draw a sketch to display the probability that one randomly selected nine-year-old boy has a height greater than the mean height for four-year-old girls. Then, use Table 2 of the Statistical Tables to find this probability.

(j)

= 132.2 = 5.7

132.2101.6 (z = –5.37)

practically one (1) or 100%

Find the probability that one randomly selected nine-year-old girl has a height within 0.5 cm. of the population mean. First, draw a sketch to display this probability; then find this probability by using Table 2 of the Statistical Tables.

5.-continued (k)

132.2

= 132.2 = 6.5

131.7

z = (131.7 – 132.2) / 6.5 = –0.08

z = (132.7 – 132.2) / 6.5 = +0.08

132.7

0.4681 0.4681

(z = –0.08)

(z = +0.08)

1 – (0.4681 + 0.4681) = 0.0638 or 6.38%

Find the probability that the mean height for a simple random sample of 25 nine-year-old girls is within 0.5 cm. of the population mean. First, obtain the mean and standard deviation for the sampling distribution and draw a sketch to display this probability; then, find this probability by using Table 2 of the Statistical Tables.

(l)

132.2

= 132.2 = 6.5

x = =

x = n =

131.7

z = (131.7 – 132.2) / 1.3 = –0.38

z = (132.7 – 132.2) / 1.3 = +0.38

132.70.3520 0.3520

(z = –0.38) (z = +0.38)

1 – (0.3520 + 0.3520) = 0.2960 or 29.60%

From the Central Limit Theorem:

132.2

6.5 25 = 1.3

Find the probability that the mean height for a simple random sample of 400 nine-year-old girls is within 0.5 cm. of the population mean. First, obtain the mean and standard deviation for the sampling distribution and draw a sketch to display this probability; then, find this probability by using Table 2 of the Statistical Tables.

5.-continued (m)5.-continued (m)

132.2

= 132.2 = 6.5

x = =

x = n =

131.7

z = (131.7 – 132.2) / 0.325 = –1.54

z = (132.7 – 132.2) / 0.325 = +1.54

132.7

0.0618 0.0618(z = –1.54) (z = +1.54)

1 – (0.0618 + 0.0618) = 0.8764 or 87.64%

From the Central Limit Theorem:

132.2

6.5 400 = 0.325

6.

(a)

(b)

The mean weight for nine-year-old girls is to be estimated with the mean weight from a simple random sample of n nine-year-old girls. Suppose we believe that the distribution of these weights is positively skewed, and that the standard deviation of these weights is about 7.0 kg.

Why can we not expect to use the normal distribution to accurately find the probability that the weight of one randomly selected nine-year-old girl is within 2 kg. of the population mean?

If the distribution of weights is positively skewed, they do not have a normal distribution.

Why might we reasonably expect to use the normal distribution to accurately find the probability that the mean weight of say n = 9 randomly selected nine-year-old girl is within 2 kg. of the population mean?

The sampling distribution of x is approximately a normal distribution with a sufficiently large sample size n.

6.-continued (c) Find the probability that the mean weight in a simple random sample of 9

nine-year-old girls will be within 2 kg. of the population mean weight. First, obtain the mean and standard deviation for the sampling distribution and draw a sketch to display this probability; then, find this probability by using Table 2 of the Statistical Tables.

–2 +2

0.1949 0.1949

(z = –0.86) (z = +0.86)

Even though we do not know , we can find the z-score corresponding to a raw score 2 kg. away from .

= ????? = 7.0

y = =

y = n =

?????

7.0 9 = 2.333

z = [(–2) – ] / 2.333 = –2 / 2.333 = –0.86

z = [(+2) – ] / 2.333 = 2 / 2.333 = +0.86

1 (0.1949 + 0.1949) = 0.6102 or 61.02%

(d) Find the probability that the mean weight in a simple random sample of 100 nine-year-old girls will be more than 2 kg. away from (in either direction) of the population mean weight. First, obtain the mean and standard deviation for the sampling distribution and draw a sketch to display this probability; then, find this probability by using Table 2 of the Statistical Tables.

–2 +2

0.4979 0.4979

(z = –2.86) (z = +2.86)

Even though we do not know , we can find the z-score corresponding to a raw score 2 kg. away from .

= ????? = 7.0

y = =

y = n =

?????

7.0 100 = 0.7

z = [(–2) – ] / 0.7 = –2 / 0.7 = –2.86

z = [(+2) – ] / 0.7 = 2 / 0.7 = +2.86

1 – (0.4979 + 0.4979) = 0.0042 or 0.42%

Before submitting Homework #2, check some of the answers (if you haven’t done so already) from the link on the course schedule:

http://srv2.lycoming.edu/~sprgene/M214/Schedule214.htm