Statistics I. International Group Departamento de Economa Aplicada Universitat de Valncia

May 20, 2010

Problem 35

Random variable X : weekly ticket sales (units) of a museum. X ∼ N(1000, 180)

Find the probability of weekly sales exceeding 850 tickets. Find the probability of the interval 1000 to 1200 Take 5 weeks at random. Find the probability of weekly sales not exceeding 850 tickets in more than two weeks Ticket price is 4.5 Euros. Deﬁne the distribution of weekly revenue

Problem 35 (a)

Random variable X : weekly ticket sales (units) of a museum. X ∼ N(1000, 180) Find the probability of weekly sales exceeding 850 tickets. Find the probability of the interval 1000 to 1200 P(X > 850)? P(X > 850) = P(Z > 850 − 1000 ) = P(Z > −0.83) = 180

P(Z < 0.83) = P(Z < 0.83) = 0.7967

Problem 35 (a)
Random variable X : weekly ticket sales (units) of a museum. X ∼ N(1000, 180) Find the probability of weekly sales exceeding 850 tickets. Find the probability of the interval 1000 to 1200 P(1000 < X < 1200)? P(1000 < X < 1200) = P(X < 1200) − P(X < 1000) = 1200 − 1000 1000 − 1000 ) − P(Z < ) 180 180 = P(Z > −0.83) = P(Z < 1.11)−P(Z < 0) = 0.8665−0.5 = P(Z < = 0.3665

Problem 35 (b)
Probability of more than two weeks with sales not exceeding 850 tickets Y : number of weeks with sales not exceeding 850 tickets in 5 weeks. The probability of sales being less than 850 tickets (found in the ﬁrst section) is 1-0.7967=0.2033. Let us analyze this random experiment Ticket sales for one week can be below or above 850 (success/failure) We repeat the previous step n = 5 trials Then the number of weeks with ticket sales below 850 in 5 trials, variable Y , follows a binomial distribution Y ∼ Bi(p = 0.2033, n = 5)

Problem 35 (b)

Probability of more than two weeks with sales not exceeding 850 tickets Y ∼ Bi(p = 0.2033, n = 5)

...NORMALDISTRIBUTION
1. Find the
distribution:
a.
b.
c.
d.
e.
f.
following probabilities, the random variable Z has standard normal
P (0< Z < 1.43)
P (0.11 < Z < 1.98)
P (-0.39 < Z < 1.22)
P (Z < 0.92)
P (Z > -1.78)
P (Z < -2.08)
2. Determine the areas under the standard normal curve between –z and +z:
♦ z = 0.5
♦ z = 2.0
Find the two values of z in standard normaldistribution so that:
P(-z < Z < +z) = 0.84
3. At a university, the average height of 500 students of a course is 1.70 m; the standard
deviation is 0.05 m. Find the probability that the height of a randomly selected student is:
1. Below 1.75 m
2. Between 1.68 m and 1.78 m
3. Above 1.60 m
4. Below 1.65m
5. Above 1.8 m
4. Suppose that IQ index follows the normaldistribution with µ = 100 and the standard
deviation σ = 16. Miss. Chi has the IQ index of 120. Find the percentage of people who
have the IQ index below that of Miss. Chi.
5. The length of steel beams made by the Smokers City Steel Company is normally
distributed with µ = 25.1 feet and σ = 0.25 feet.
a. What is the probability that a steel beam will be less than 24.8 feet long?
b. What is the probability that a steel beam will be more than 25.25 feet
long?
c. What is the probability that a steel beam will be between 24.9 and 25.7
feet long?
d. What is the probability that a steel beam will be between 24.6 and 24.9
feet long?
e....

...require that we know whether we have a sample or a population. 2. The following numbers represent the weights in pounds of six 7year old children in Mrs. Jones' 2nd grade class. {25, 60, 51, 47, 49, 45} Find the mean; median; mode; range; quartiles; variance; standard deviation. Solution: mean = 46.166.... median = 48 mode does not exist range = 35 Q1 = 45 Q2 = median = 48 Q3 = 51 variance = 112.1396 standard deviation =10.59 3. If the variance is 846, what is the standard deviation? Solution: standard deviation = square root of variance = sqrt(846) = 29.086 4. If we have the following data
34, 38, 22, 21, 29, 37, 40, 41, 22, 20, 49, 47, 20, 31, 34, 66 Draw a stem and leaf. Discuss the shape of the distribution. Solution: 2 3 4 5 6 | | | | | 219200 48714 0197 6
This distribution is right skewed (positively skewed) because the “tail” extends to the right. 5. What type of relationship is shown by this scatter plot?
45 40 35 30 25 20 15 10 5 0 0 5 10 15 20
Solution: Weak positive linear correlation 6. What values can r take in linear regression? Select 4 values in this interval and describe how they would be interpreted. Solution: the values are between –1 and +1 inclusive. -1 means strong negative correlation +1 means strong positive correlation 0 means no correlation .5 means moderate positive correlation etc. 7. Does correlation imply causation? Solution: No.
8. What do we call the r value. Solution: The correlation coefficient....

...6
The NormalDistribution
Objectives
Outline
After completing this chapter, you should be able to
1
2
3
Identify distributions as symmetric or skewed.
4
Find probabilities for a normally distributed
variable by transforming it into a standard
normal variable.
Introduction
6–1
NormalDistributions
Identify the properties of a normaldistribution.
Find the area under the standard normaldistribution, given various z values.
5
Find speciﬁc data values for given
percentages, using the standard normaldistribution.
6
6–3 The Central Limit Theorem
6–4 The Normal Approximation to the Binomial
Distribution
Use the central limit theorem to solve
problems involving sample means for large
samples.
7
6–2 Applications of the NormalDistribution
Use the normal approximation to compute
probabilities for a binomial variable.
Summary
6–1
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Confirming Pages
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Chapter 6 The NormalDistribution
Statistics
Today
What Is Normal?
Medical researchers have determined so-called normal intervals for a person’s blood
pressure, cholesterol,...

...under a Standard Normal curve
a) to the right of z is 0.3632;
b) to the left of z is 0.1131;
c) between 0 and z, with z > 0, is 0.4838;
d) between -z and z, with z > 0, is 0.9500.
Ans : a) z = + 0.35 ( find 0.5- 0.3632 = 0.1368 in the normal table)
b) z = -1.21 ( find 0.5 – 0.1131 = 0.3869 in the normal table)
c ) the area between 0 to z is 0.4838, z = 2.14
d) the area to the right of +z = ( 1-0.95)/2 = 0.025, therefore z = 1.96
3. Given the Normally distributed variable X with mean 18 and standard deviation 2.5, find
a) P(X < 15);
b) the value of k such that P(X < k) = 0.2236;
c) the value of k such that P(X > k) = 0.1814;
d) P( 17 < X < 21).
Ans : X ~ N ( 18, 2.52)
a) P ( X < 15)
P ( Z < (15-18)/2.5) = P ( Z < -1.2) = 0.1151 ( 4 decimal places)
b) P ( X < k) = 0.2236
P ( Z < ( k – 18) / 2.5 ) = 0.2236
From normal table, 0.2236 = -0.76
(k-18)/2.5 = - 0.76, solve k = 16.1
c) P (X > k) = 0.1814
P ( Z > (k-18)/2.5 ) = 0.1814
From normal table, 0.1814 = 0.91
(k-18)/ 2.5 = 0.91, solve k = 20.275
d) P ( 17 < X < 21)
P ( (17 -18)/2.5 < Z < ( 21-18)/2.5)
P ( -0.4 < Z < 1.2) = 0.8849 – 0.3446 = 0.5403 ( 4 decimal places)...

...
NormalDistribution
Student’s Name:
Instructorr:
Date ofSubmission:
The table 1 below shows a relationship between actual daily temperatures and precipitation in the month of January 2011. These data was adopted from a meterological station in the states of Alaska, in the United States. Normal distributed data aresymmetric with a single bell shaped peaks. Th maean of the data it significant in indication the point that the peak is likely to occur. In addition, standard deviation indicates the spread, which is usually referred to asthegirth of thebeell shapedcurve (Balakrishnan & Nevzorov, 2003).
Table 1: Relationship between actual daily temperatures and precipitation of January 2011
|Date |Actual Temperatures |Precipitation |
|Jan. 1 |30 |0 |
|Jan. 2 |25 |0 |
|Jan. 3 |31 |0 |
|Jan. 4 |33 |0 |
|Jan. 5 |29 |0 |
|Jan. 6 |36 |0.26 |
|Jan. 7 |36 |0 |
|Jan. 8 |37 |0.01 |
|Jan. 9 |32 |0.21 |
|Jan. 10 |28...

...should really do to put a stop to the LRA, and I don’t think it’s exactly to purchase merchandise from an organization whose top three representatives received over $80,000 each annually, solely from donations. That’s over $200,000 not including expenses back and forth from Africa and a minimum of housing and schools being built.
If you look on IC's official website, the only infrastructure they have built outside of Uganda since the LRA left Uganda is a small rehabilitation center that can house 200 or so people. The rehabilitation center is a nice thought, but the fact that a 13 million dollar a year revenue charity can only build one rehabilitation center in the 7 years that the LRA has moved to South Sudan just highlights the inherent problem of IC being based in Uganda.
Works Cited
1. Josiah. "Concerning KONY 2012 - Nothing Suspicious Going On Here." Nothing Suspicious Going On Here. 7 Mar. 2012. Web. 13 Mar. 2012. .
2. Manhire, Toby. "Kony 2012: Some Reading." â The Internaut â Commentary â New Zealand Listener. 12 Mar. 2012. Web. 13 Mar. 2012. .
3. "Kony 2012." Invisible Children. Ed. Ben Keesey. 7 Mar. 2012. Web. 13 Mar. 2012. .
4. Russell, Jason. "Kony 2012." YouTube. YouTube, 7 Mar. 2012. Web. 25 Mar. 2012. .
...

...Dropbox Quick Start
What is Dropbox?
Dropbox is a free service that lets you bring all your photos, docs, and videos anywhere. Any file you
save to your Dropbox will also automatically save to all your computers, phones, and even the Dropbox
website. This means that you can start working on your computer at school or the office, and finish on
your home computer. Never email yourself a file again!
The Dropbox Folder
After you install Dropbox on your computer, a Dropbox folder is created. If you’re reading this guide,
then that means you had no problems finding your Dropbox folder :). This folder is just like any other
folder on your computer, but with a twist. Any file you save to your Dropbox folder is also saved to all
your other computers, phones, and the Dropbox website.
On top of your Dropbox is a green icon that lets you know how your Dropbox is doing:
Green circle and check: All the files in your Dropbox are up to date.
Blue circle and arrows: Files in your Dropbox are currently being updated.
Adding files to your Dropbox
Step 1 Drag and drop a file into your Dropbox folder.
Step 2 The blue icon means your file’s syncing with Dropbox.
That’s it! The green icon means that your ﬁle has finished saving
to your other computers and the Dropbox website. Now that your
file’s in Dropbox, any changes made to it will be automatically
detected and updated to your other computers.
The Dropbox Right-click/Context Menu
When you Right-Click...

...range = 60
mode = 165
variance = 324
median = 170
The coefficient of variation equals
a. 0.1125%
b. 11.25%
c. 203.12%
d. 0.20312%
____3. In a binomial experiment, which one(s) of the following is (are) true?
(i) The probability of success in the second trial is dependent on the outcome in the first trial.
(ii) Only two outcomes are possible in each trial.
(iii) The probability of success in each trial is always equal to the probability of failure.
(iv) The expected value is always greater than or equal to the variance.
a. (ii) only
b. (ii) and (iv)
c. (iii) and (iv)
d. (i), (ii) and (iv).
____ 4. A normaldistribution with a mean of 0 and a standard deviation of 1 is called
a. a probability density function
b. an ordinary normal curve
c. a standard normaldistribution
d. none of these alternatives is correct
Exhibit 1
The student body of a large university consists of 60% female students. A random sample of 8 students is selected.
____ 5. Refer to Exhibit 1. What is the probability that among the students in the sample at least 7 are female?
a. 0.1064
b. 0.0896
c. 0.0168
d. 0.8936
Exhibit 2
The average price of personal computers manufactured by MNM Company is $1,200 with a standard deviation of $220. Furthermore, it is known that the computer prices manufactured by MNM are normally distributed.
____6. Refer to...