Introduction to Business Statistics
7th Edition
ISBN: 9781111792374
Author: WEIERS
Publisher: Cengage Learning
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Chapter 2.2, Problem 2.1E
What is a frequency distribution? What benefits does it offer in the summarization and reporting of data values?
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How many apps? According to a website, the mean number of apps on a smartphone in the United States is 82. Assume
the number of apps is normally distributed with mean 82 and standard deviation 30.
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(a) What proportion of phones have more than 47 apps? Round the answer to four decimal places.
The proportion of phones that have more than 47 apps is 0.8783
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You are provided with data that includes all 50 states of the United States. Your task is to draw a sample of:
20 States using Random Sampling (2 points: 1 for random number generation; 1 for random sample)
10 States using Systematic Sampling (4 points: 1 for random numbers generation; 1 for generating random sample different from the previous answer; 1 for correct K value calculation table; 1 for correct sample drawn by using systematic sampling) (For systematic sampling, do not use the original data directly. Instead, first randomize the data, and then use the randomized dataset to draw your sample. Furthermore, do not use the random list previously generated, instead, generate a new random sample for this part. For more details, please see the snapshot provided at the end.) You are provided with data that includes all 50 states of the United States. Your task is to draw a sample of: o 20 States using Random Sampling (2 points: 1 for random number generation; 1 for random sample) o…
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Listed in the accompanying table are weights (kg) of randomly selected U.S. Army male personnel measured in 1988 (from "ANSUR I 1988") and different weights (kg) of randomly selected U.S.
Army male personnel measured in 2012 (from "ANSUR II 2012"). Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not
assume that the population standard deviations are equal. Complete parts (a) and (b).
Click the icon to view the ANSUR data.
a. Use a 0.05 significance level to test the claim that the mean weight of the 1988…
Chapter 2 Solutions
Introduction to Business Statistics
Ch. 2.2 - What is a frequency distribution? What benefits...Ch. 2.2 - Prob. 2.2ECh. 2.2 - The National Safety Council reports the following...Ch. 2.2 - Prob. 2.4ECh. 2.2 - Prob. 2.5ECh. 2.2 - Using the frequency distribution in Exercisc 2.5,...Ch. 2.2 - What is meant by the statement that the set of...Ch. 2.2 - For commercial banks in each state, the U.S....Ch. 2.2 - The accompanying data describe the hourly wage...Ch. 2.2 - The following performance scores have been...
Ch. 2.2 - During his career in the NHL, hockey great Wayne...Ch. 2.2 - According to the U.S. Department of Agriculture,...Ch. 2.2 - Convert the distribution in Exercise 2.3 to a...Ch. 2.2 - Prob. 2.14ECh. 2.2 - Using the frequency distribution obtained in...Ch. 2.2 - For the frequency distribution constructed in...Ch. 2.2 - Prob. 2.17ECh. 2.2 - Prob. 2.18ECh. 2.2 - Prob. 2.19ECh. 2.3 - Construct a stem-and-leaf display for the...Ch. 2.3 - Prob. 2.21ECh. 2.3 - In the following stem-and-leaf display for a set...Ch. 2.3 - Prob. 2.23ECh. 2.3 - Prob. 2.24ECh. 2.3 - Prob. 2.25ECh. 2.3 - Prob. 2.26ECh. 2.3 - Prob. 2.28ECh. 2.4 - What is the difference between a histogram and a...Ch. 2.4 - Prob. 2.30ECh. 2.4 - Prob. 2.31ECh. 2.4 - Prob. 2.32ECh. 2.4 - Prob. 2.33ECh. 2.4 - Prob. 2.34ECh. 2.4 - It has been estimated that 92.9% of U.S....Ch. 2.4 - Prob. 2.37ECh. 2.5 - What is a scatterplot, and for what kind of data...Ch. 2.5 - Prob. 2.39ECh. 2.5 - Prob. 2.40ECh. 2.5 - Prob. 2.41ECh. 2.5 - For six local offices of a large tax prepartion...Ch. 2.5 - In the 2009 stress tests applied to the nation’s...Ch. 2.5 - Prob. 2.44ECh. 2.5 - Prob. 2.45ECh. 2.6 - Prob. 2.46ECh. 2.6 - Difference between simple tabulation and...Ch. 2.6 - Prob. 2.48ECh. 2.6 - For the fleet described in Exercise 2.48, a....Ch. 2.6 - Prob. 2.50ECh. 2.6 - Prob. 2.51ECh. 2.6 - Prob. 2.52ECh. 2.6 - Prob. 2.53ECh. 2.6 - Prob. 2.54ECh. 2 - The breakdown of U.S. cities having a population...Ch. 2 - The National Oceanic and Atmospheric...Ch. 2 - Prob. 2.57CECh. 2 - Prob. 2.58CECh. 2 - Prob. 2.59CECh. 2 - Prob. 2.60CECh. 2 - The following stem-and-leaf output has been...Ch. 2 - For the period 2001—2008, the lristol-Myers Squibb...Ch. 2 - Prob. 2.63CECh. 2 - Prob. 2.64CECh. 2 - Prob. 2.65CECh. 2 - Prob. 2.66CECh. 2 - Prob. 2.67CECh. 2 - Prob. 2.68CECh. 2 - Prob. 2.69CECh. 2 - Prob. 2.70CECh. 2 - Using 10-yard intervals beginning with 200.0—under...Ch. 2 - Using the same intervals as in part 1, construct a...Ch. 2 - Prob. 1.3IC
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- solving problem 1arrow_forwardselect bmw stock. you can assume the price of the stockarrow_forwardThis problem is based on the fundamental option pricing formula for the continuous-time model developed in class, namely the value at time 0 of an option with maturity T and payoff F is given by: We consider the two options below: Fo= -rT = e Eq[F]. 1 A. An option with which you must buy a share of stock at expiration T = 1 for strike price K = So. B. An option with which you must buy a share of stock at expiration T = 1 for strike price K given by T K = T St dt. (Note that both options can have negative payoffs.) We use the continuous-time Black- Scholes model to price these options. Assume that the interest rate on the money market is r. (a) Using the fundamental option pricing formula, find the price of option A. (Hint: use the martingale properties developed in the lectures for the stock price process in order to calculate the expectations.) (b) Using the fundamental option pricing formula, find the price of option B. (c) Assuming the interest rate is very small (r ~0), use Taylor…arrow_forward
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- 0|0|0|0 - Consider the time series X₁ and Y₁ = (I – B)² (I – B³)Xt. What transformations were performed on Xt to obtain Yt? seasonal difference of order 2 simple difference of order 5 seasonal difference of order 1 seasonal difference of order 5 simple difference of order 2arrow_forwardCalculate the 90% confidence interval for the population mean difference using the data in the attached image. I need to see where I went wrong.arrow_forwardMicrosoft Excel snapshot for random sampling: Also note the formula used for the last column 02 x✓ fx =INDEX(5852:58551, RANK(C2, $C$2:$C$51)) A B 1 No. States 2 1 ALABAMA Rand No. 0.925957526 3 2 ALASKA 0.372999976 4 3 ARIZONA 0.941323044 5 4 ARKANSAS 0.071266381 Random Sample CALIFORNIA NORTH CAROLINA ARKANSAS WASHINGTON G7 Microsoft Excel snapshot for systematic sampling: xfx INDEX(SD52:50551, F7) A B E F G 1 No. States Rand No. Random Sample population 50 2 1 ALABAMA 0.5296685 NEW HAMPSHIRE sample 10 3 2 ALASKA 0.4493186 OKLAHOMA k 5 4 3 ARIZONA 0.707914 KANSAS 5 4 ARKANSAS 0.4831379 NORTH DAKOTA 6 5 CALIFORNIA 0.7277162 INDIANA Random Sample Sample Name 7 6 COLORADO 0.5865002 MISSISSIPPI 8 7:ONNECTICU 0.7640596 ILLINOIS 9 8 DELAWARE 0.5783029 MISSOURI 525 10 15 INDIANA MARYLAND COLORADOarrow_forward
- Suppose the Internal Revenue Service reported that the mean tax refund for the year 2022 was $3401. Assume the standard deviation is $82.5 and that the amounts refunded follow a normal probability distribution. Solve the following three parts? (For the answer to question 14, 15, and 16, start with making a bell curve. Identify on the bell curve where is mean, X, and area(s) to be determined. 1.What percent of the refunds are more than $3,500? 2. What percent of the refunds are more than $3500 but less than $3579? 3. What percent of the refunds are more than $3325 but less than $3579?arrow_forwardA normal distribution has a mean of 50 and a standard deviation of 4. Solve the following three parts? 1. Compute the probability of a value between 44.0 and 55.0. (The question requires finding probability value between 44 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 44, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the answer of the second part.) 2. Compute the probability of a value greater than 55.0. Use the same formula, x=55 and subtract the answer from 1. 3. Compute the probability of a value between 52.0 and 55.0. (The question requires finding probability value between 52 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 52, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the…arrow_forwardIf a uniform distribution is defined over the interval from 6 to 10, then answer the followings: What is the mean of this uniform distribution? Show that the probability of any value between 6 and 10 is equal to 1.0 Find the probability of a value more than 7. Find the probability of a value between 7 and 9. The closing price of Schnur Sporting Goods Inc. common stock is uniformly distributed between $20 and $30 per share. What is the probability that the stock price will be: More than $27? Less than or equal to $24? The April rainfall in Flagstaff, Arizona, follows a uniform distribution between 0.5 and 3.00 inches. What is the mean amount of rainfall for the month? What is the probability of less than an inch of rain for the month? What is the probability of exactly 1.00 inch of rain? What is the probability of more than 1.50 inches of rain for the month? The best way to solve this problem is begin by a step by step creating a chart. Clearly mark the range, identifying the…arrow_forward
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