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Toronto Metropolitan University QMS 210 Parameter Vs Statistic -Statistic is a sample of a population - Parameter is a whole population Example 1: The average score for a class of 28 students taking a calculus midterm exam was 72%. Answer: The value is a parameter because the class of 28 students is a population Example 2: Some peo ...[Show More]

Toronto Metropolitan University
QMS 210
Parameter Vs Statistic

-Statistic is a sample of a population

- Parameter is a whole population

Example 1: The average score for a class of 28 students taking a calculus midterm exam was 72%.

Answer: The value is a parameter because the class of 28 students is a population

Example 2: Some people responded to this request: "Dial 1-900-PRO-LIFE to participate in a telephone poll

on abortion. ($4.45 per minute. Average call: 2 minutes. You must be 18 years old.)"

a) Identify the sample b) Identify the population. c) Determine whether the sample is likely to be

representative of the population.

Answer: a) The sample is all individuals who responded. b) The population is all individuals who are at least

18 years of age. c) No, the sample is not likely to be representative of the population because those with

strong opinions about abortion are more likely to respond.

Categorical (qualitative) or Numerical

Example: a) Expected time until return b) Weight c) Academic minor d) Number of computers owned e) The

number of stores a person visited in the past month f) A person's favorite department store f) The time a

person spent exercising in the past month

Answer: a) Numerical Continuous (Ratio) b) Numerical Continuous (Ratio) c) Qualitative (Nominal) d)

Numerical Discrete (Ratio) e) Discrete Numerical (Ratio) f) Qualitative (Nominal) g) Continuous Numerical

(Ratio)

Stem & Leaf Plot

Ordered Array+ Stem & Leaf Plot

-Organize from lowest number to highest number

Example: 1.9, 3.2, 5.8, 7.6, 8.6, 9.4, 9.6, 9.9, 10.1, 10.2, 10.3, 11.1, 11.4, 12.4, 13.4, 14.3, 14.7, 16.1, 17.3,

26.8, 32.2, 32.4, 70.1, 92.5, 113.1, 127.2 (round to nearest integer, 1-10 is in 0 row)

0 236899

1 0000011234567

2 7

3 22

456

7 0

8

9 3

10

11 3

12 7

RF & CRF

Example: A Canadian generic pharmaceutical manufacturer sells 78 different products. The distribution of

the annual sales amounts last year for these products is shown below.

Answer: RF= F/N and CFR= accumulation starting from first number (N= 78) (F= 17)

Social Recommendations

Example: Those who arrived at the link from a recommendation had a correct recall of 62.07% as compared

to those who arrived at the link from browsing who had a correct recall of 62.91%. Since the difference in

percentage is relatively small, one can conclude that social recommendations have very little impact on

correct recall.

16. A survey of 2350 adults reported that 51% watch news videos.

Complete parts (a) through (c) below. Suppose that you take a sample of

50 adults. If the population proportion of adults who watch news videos is

0.51, what is the probability that fewer than half in your sample will watch

news videos? N=50, Po=0.51, Mean of sampling distribution (µ ) = Po = 0.51

Step 1: Find population variance √Po∗(1n-Po)

Step 2: Find Z-score = P-µ/ σ =

0.5-0.51/ 0.0707 = -0.14  0.4443 is probability

17. The distribution of weekly salaries at a large company is reverse J-shaped

with a mean of $1000 and a standard deviation of $370. What is the

probability that the sampling error made in estimating the mean weekly

salary for all employees of the company by the mean of a random sample of

weekly salaries of 80 employees will be at most $75? N= 80, SD= 370, = 75

 √80 = [z*370/75]^2  √80 = 4.93z  z = 1.81

7. How to find percentage of trucks can be expected to travel either less than 35 or more than 70 thousand

miles in a year? Use Normal C.D. on calculator, Lower=35, Upper=70, μ=50, σ=11, 1-P= ANSWER

c) How many miles will be traveled by at least 60% of the trucks? Use Inverse Normal on calculator, Tail= Right

because above 50%, Area=0.6, μ=50, σ=11

8. Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1), complete

parts (a) through (d).

a) What is the probability that Z < 1.58? Use Inverse Normal, Tail Left, plug in multiple choice (IF Z<, then tail

left, IF Z>, then tail is right)

b) What is the probability that Z > 1.87? Plug into Normal C.D. as Lower: 1.87, Upper: 100, μ=0, σ=1

c) What is the probability that 1.58 < Z > 1.87? Use Calculator for this

d) What is the value of Z if only 11.51% of all possible Z-values are larger? Use Inverse Normal, Tail: Left, Area:

0.8849, μ=0, σ=1

9. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard

deviation of 0.8 pound. A citation catfish should be one of the top 2% in weight. Assuming the weights of

catfish are normally distributed, at what weight (in pounds) should the citation designation be established?

Use Inverse Normal on calculator, Tail= Left, Area=0.98, μ=3.2, σ=0.8

10. The probability that Z<1.55 or Z>1.83. It has a mean of o and SD of 1. Use Normal C.D. on calculator,

Lower=1.55, Upper=1.83, μ=0, σ=1, 1-P= ANSWER

The Central Limit Theorem = is important in statistics because for a large n, it says the sampling distribution of

the sample mean is approximately normal, regardless of the shape of the population.

11. When probability is 55% that the sample mean will be between what two values symmetrically

distributed around the population mean? Step 1: 0.55 x u. Step 2: step 1 answer x σ/ √n . Step: u-step

2 answer< u < 1.31+step 2 answer

NOTE: For sample sizes greater than 30, the sampling distribution of the mean will be approximately normally

distributed regardless of the shape of the population. Less it will be symmetrical

SAMPLE MEANS sample means between two values

Step 1: Equation: (n-μ) / (σ / √n ) = P(X<4) = (4-3.2) / (0.8/ Sqrt (4)) = 0.8/0.4 = 2 --- 0.9772

P (X>3) = (3-3.2) / (0.8/ Sqrt ( 4 )) = -0.2/0.4 = -0.50 --- 0.3085

Step 2: 0.9772 - 0.3085 = 0.6687 (67%)

13. Given a normal distribution with μ = 50 and σ=5 and given you select a sample of n equals 100

Equation: (n-μ) / (σ / √n )

a) What is the probability that Upper X is less than 49? P(X<49) = (49-50) / (5/ Sqrt

(100)) = -1/0.5 = -2 --- 0.0228

b) There is a 30% chance that Upper X overbar is above what value?

Step 1 = 0.3000 = 0.7000. Go on positive chart and look for

closest value to 0.7000. It is 0.53 Step 2: Equation: μ + z x (σ / √n ) = 50+0.53 x

(5/10) = 50+ (0.53 x 0.5) = 50.265 or 50.27

Find Random Sample mean and SD = Mean (keep it)  20 and SD  SD/sqrt (sample)16/ √64 = 2

a) Calculate the standard normal z-score corresponding to a value of x = 16. z= (x-u)/ σx (σx = 2)

b) Find P(x < 16) and P(x > 16) P(x < 16) = μ + z x (σ / √n ) AND P(x > 16) = do 1-P because GREATER

Degree of Freedom: # of samples there are - 1

If distribution is J-shaped. Given mean, SD, and sample and at most $75 (E).  Find Z n= (Z x σ / E

(error))^2

If normally distributed and at most 0.02  sample – mean / SD  take –z and +z  minus each other

15. Suppose the life of a particular brand of calculator battery is approximately normally distributed with a

mean of 80 hours and a standard deviation of 10 hours.

a) What is the probability that a single battery randomly selected from the population will have a life between

75 and 85 hours? Plug into Normal C.D. as Lower: 75, Upper: 75, μ=80, σ=10

b) What is the probability that 9 randomly sampled batteries from the population will have a sample mean life

1. According to an airline, flights on a certain route are on time 80% of the time. Suppose 25 flights are

randomly selected and the number of on-time flights is recorded.

A) Find and interpret the probability that exactly 15 flights are on time. Use Binomial P.D. on calculator

X=15 flight, Numtrial= 25 flight, P=0.8

(B) Find and interpret the probability that fewer than 14 flights are on time. Use Binomial C.D. on calculator

X=15 flight, Numtrial= 25 flight, P=0.8

(C) Find and interpret the probability that at least 15 flights are on time. Use Binomial C.D. on calculator X=14

flight, Numtrial= 25 flight, P=0.8, DO 1-P= answer

(D) Find and interpret the probability that between 13 and 15 flights, inclusive, are on time. Use Binomial C.D.

on calculator . Use X-1 for lowest value, so 12 and 15 flights. Minus each other to get answer.

(E) Find the probability of 15 or more successes Use Binomial C.D. on calculator X=14 flight, Numtrial= 25

flight, P=0.8, DO 1-P= answer

(E) Find the probability of 15 or fewer successes Use Binomial C.D. on calculator X=15 flight, Numtrial= 25

flight, P=0.8

If question like exceeded $, do normal C.D. Find S.D. by SD/sqrt (sample)

Poisson Distribution  (if question like with a mean equal to 10 per hour. What are the chances that no

customers will arrive in a 15-minute period?  15/60mins  0.25 x 10  Poisson PD u = 2.5 and x=1-

Example: Assume a Poisson distribution with λ=5.4

P(X=1)= ? Use Poisson P.D. on calculator , X=1, µ=5.4

P(X<1)= ? Use Poisson C.D. on calculator, X=0, µ=5.4

P(X>1)= ? 1-P(X≤1) = answer

P(X≤1)= ? Use Poisson C.D. on calculator, X=1, µ=5.4

2. The probability that a call received by a certain switchboard will be a wrong number is 0.02. Use the

Poisson distribution to approximate the probability that among 150 calls received by the switchboard, there

are at least two wrong numbers. Round your answer to four decimal places.

Mean=150*0.02=3  1-e^-3 (1+3) = 0.8009

3. The probability is 0.00005. Use the Poisson distribution to approximate the probability that among 14,000

objects, exactly 2 will have a problem. Round your answer to four decimal places. (14,000 x 0.00005) = 0.7

4. As survey found that women's heights are normally distributed with am mean 62.1 in. and standard

deviation 2.9. The survey also found that men's heights are normally distributed with mean 67.8 and

standard deviation 3.1 in. Consider an executive jet that seats six with a doorway height of 55.8 in.

a) what percentage of adult men can fit through the door without bending?

Step 1: Find Z score

Use Normal C.D. on calculator, Upper=55.8, Lower= 0, μ=67.8, σ=3.1, Answer is z: Up

b) Does the door design with a height of 56.1 in. appear to be adequate? Why didn't the engineers design a

larger door? The door design is inadequate, but because the jet is relatively small and seats only six people, a

much higher door would require major changes in the design and cost of thejet, making a larger height not

practical.

c) What doorway height would allow 40% of men to fit without bending? Use Inverse Normal on calculator,

Tail: Left, Area: 0.4, μ=67.8, σ=3.1

5. Given a normal distribution with μ=100 and σ=10

a) The probability that X>75 is: Use Normal C.D. on calculator, X=75, Upper=200, Lower= 75, μ=100, σ=10