Download Solved Problems on Statistical Methods in Normal Distribution | STA 2023 and more Study notes Data Analysis & Statistical Methods in PDF only on Docsity!
Practice Demo: Suppose that 33 percent of women believe in the existence of aliens. If 100 women are selected at random, what is the probability that more than 45 percent of them will say that they believe in aliens?
SET UP:
Role #1: “100 women selected” “45 percent of them”
Role #2: p ˆ^ μ( p ˆ ) = p ( )
n
p p SD p
ˆ (^1 − )
Role #3: μ( p ˆ) =0.33 ( ) ≈
ˆ^0.^33 (^10.^33 )
SD p 0.
Role #4:
p ˆ
.19 .24 .28 .33 .38 .42.
SOLUTION:
p ˆ
.19 .24 .28 .33 .38 .42.
normalcdf( .45, 1E99, 0.33, 0.047 ) = 0.
- Suppose family incomes in a town are normally distributed with a mean of $1, and a standard deviation of $600 per month. What is the probability that a family has an income between $1,400 and $2,250?
SET UP:
Role #1: No sample of size greater than one was taken. Family incomes are the population. Role #2: X μ σ Role #3: μ = 1200 σ = 600 Role #4:
X
SOLUTION:
X
normalcdf( 1400, 2250, 1200, 600 ) = 0.
- Find the area under the curve between the z-scores of -2 and 1.
SET UP:
Role #1: No sample of size greater than one was taken. “z-scores” Role #2: z μ = 0 σ= 1 Role #3: μ = 0 σ = 1 Role #4:
Z
-3 -2 -1 0 1 2 3
SOLUTION:
Z
-3 -2 -1 0 1 2 3
normalcdf( -2, 1, 0, 1 ) = 0.
- Adult nose length is normally distributed with mean 45mm and standard deviation 6mm. Find the probability that the sample mean nose length is between 44mm and 46mm for random samples of 36 adults.
SET UP:
Role #1: “samples of 36 adults” “ sample mean nose length”
Role #2: x μ( x )= μ ( )
n
SDx
σ
Role #3: μ( x ) = 45 ( ) = =
SD x 1
Role #4:
x
42 43 44 45 46 47 48
SOLUTION:
x
42 43 44 45 46 47 48
By the Empirical Rule we see that the answer is about 68% because 44mm and 46mm is exactly one standard deviation each way on the sample mean nose length distribution ( x distribution). More precisely we have
normalcdf( 44, 46, 45, 1 ) = 0.68269 or 68.269%
- A restaurateur anticipates serving about 180 people on a Friday evening, and believes that about 20% of the patrons will order the chef’s steak special. How many of those meals should he plan on serving in order to be pretty sure of having enough steaks on hand to meet customer demand? Justify your answer, including an explanation of what “pretty sure” means to you.
SET UP:
Role #1: “serving about 180 people” “20% of the patrons”
Role #2: p ˆ^ μ( p ˆ ) = p ( )
n
p p SD p
Role #3: μ( p ˆ) =0.2 ( ) ≈
SD p ˆ 0.
Role #4:
p ˆ
.11 .14 .17 .2 .23 .26.
SOLUTION:
Here the population is all patrons that eat at that particular restaurant on Friday nights. The 180 people on this Friday evening is a sample (although not SRS!). The proportion of those 180 people ordering the chef’s steak special is the sample proportion or p ˆ value. What could this value be? According to the Empirical Rule, we know that about 99.7% of all p ˆ^ values occur between 0.11 and 0.29. It is highly unlikely that p ˆ^ is greater than 0.29 since this happens only about 0.15% of the time (^0.^3 % 2 = 0. 15 %). Therefore, we would expect that the proportion of the 180 patrons that order the chef’s steak would be no more than 0.29. Since 29% of 180 people is 52.2 people, we conclude that the restaurateur should plan on serving 53 of those meals. That way the restaurateur can be “pretty sure” that orders of chef’s steak on Friday evenings can be filled (about 99.7% of Friday evenings).
- In this example we will be interested in the heights of northern European males. We take such a person and reduce him to a single number via the usual operations for measuring someone's height. Then we model the height of northern European males as a normal population with mu = 150 cm and sigma = 30 cm. If we sample one northern European male, what's the probability that his height will fall outside of 140 and 170? In other words, what are the chances that he'll be either below 140, or he'll be above 170 in height? That's what we mean by the word "outside."
SET UP:
Role #1: “sample one northern European” Heights of northern European males are the population. Role #2: X μ σ Role #3: μ = 150 σ = 30 Role #4:
X
60 90 120 150 180 210 240
SOLUTION:
X
60 90 120 150 180 210 240
normalcdf( -1E99, 140, 150, 30 ) = 0. normalcdf( 170, 1E99, 150, 30 ) = 0. The probability that his height will fall outside 140 cm and 170 cm is 0.36944 + 0.25249 = 0.62193 or about 62.2% of the time.
