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Hello! I need help explaining the whole procedure, the meaning of the graphs and theorems (images) and how the final result was obtained Thanks

Hello! I need help explaining the whole procedure, the meaning of the graphs and theorems (images) and how the final result was obtained Thanks

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Hello! I need help explaining the whole procedure, the meaning of the graphs and theorems (images) and how the final result was obtained

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FIGURE 7.4 Sampling distribution of for n = 2 dice FIGURE 7.5 MINITAB sampling distri- bution of x for n = 3 dice p(x-bar) .15- p(x-bar) .10- .05- 0 Using MINITAB, we generated the sampling distributions of when n = 3 and n = 4. For n = 3, the sampling distribution in Figure 7.5 clearly shows the mound shape of the normal probability distribution, still centered at μ = 3.5. Notice also that the spread of the distribution is slowly decreasing as the sample size n increases. Figure 7.6 dramatically shows that the distribution of is approximately normally distributed based on a sample as small as n = 4. This phenomenon is the result of an important statistical theorem called the Central Limit Theorem (CLT). 15- .10- .05- 0 Average of Two Dice 2 4 Average of Three Dice 5 x-bar 6 x-bar
Transcribed Image Text:FIGURE 7.4 Sampling distribution of for n = 2 dice FIGURE 7.5 MINITAB sampling distri- bution of x for n = 3 dice p(x-bar) .15- p(x-bar) .10- .05- 0 Using MINITAB, we generated the sampling distributions of when n = 3 and n = 4. For n = 3, the sampling distribution in Figure 7.5 clearly shows the mound shape of the normal probability distribution, still centered at μ = 3.5. Notice also that the spread of the distribution is slowly decreasing as the sample size n increases. Figure 7.6 dramatically shows that the distribution of is approximately normally distributed based on a sample as small as n = 4. This phenomenon is the result of an important statistical theorem called the Central Limit Theorem (CLT). 15- .10- .05- 0 Average of Two Dice 2 4 Average of Three Dice 5 x-bar 6 x-bar
1.4 FIGURE 7.3 Probability distribution for x, the number appearing on a single toss of a die THE CENTRAL LIMIT THEOREM The Central Limit Theorem states that, under rather general conditions, sums and means of random samples of measurements drawn from a population tend to have an approximately normal distribution. Suppose you toss a balanced die n = 1 time. The random variable x is the number observed on the upper face. This familiar random variable can take six values, each with probability 1/6, and its probability distribution is shown in Figure 7.3. The shape of the distribution is flat or uniform and symmet- ric about the mean = 3.5, with a standard deviation = 1.71. (See Section 4.8 and Exercise 4.84.) p(x) 1/6- 0 2 x Now, take a sample of size n = 2 from this population; that is, toss two dice and record the sum of the numbers on the two upper faces, Ex; = x₁ + x₂. Table 7.5 shows the 36 possible outcomes, each with probability 1/36. The sums are tabulated, and each of the possible sums is divided by n = 2 to obtain an average. The result is the sampling distribution of x = Ex/n, shown in Figure 7.4. You should notice the dra- matic difference in the shape of the sampling distribution. It is now roughly mound- shaped but still symmetric about the mean μ = 3.5.
Transcribed Image Text:1.4 FIGURE 7.3 Probability distribution for x, the number appearing on a single toss of a die THE CENTRAL LIMIT THEOREM The Central Limit Theorem states that, under rather general conditions, sums and means of random samples of measurements drawn from a population tend to have an approximately normal distribution. Suppose you toss a balanced die n = 1 time. The random variable x is the number observed on the upper face. This familiar random variable can take six values, each with probability 1/6, and its probability distribution is shown in Figure 7.3. The shape of the distribution is flat or uniform and symmet- ric about the mean = 3.5, with a standard deviation = 1.71. (See Section 4.8 and Exercise 4.84.) p(x) 1/6- 0 2 x Now, take a sample of size n = 2 from this population; that is, toss two dice and record the sum of the numbers on the two upper faces, Ex; = x₁ + x₂. Table 7.5 shows the 36 possible outcomes, each with probability 1/36. The sums are tabulated, and each of the possible sums is divided by n = 2 to obtain an average. The result is the sampling distribution of x = Ex/n, shown in Figure 7.4. You should notice the dra- matic difference in the shape of the sampling distribution. It is now roughly mound- shaped but still symmetric about the mean μ = 3.5.
Expert Solution
Step 1

If n = 1 a single dice

If a single dice is thrown the probability of getting {1,2,3,4,5,6} = 1/6

Each value is having equal chance of occurrence then the probability is 1/6

x p(x)
1 1/6
2 1/6
3 1/6
4 1/6
5 1/6
6 1/6
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Hi! from the previous explanation I have doubts about how the probabilities of the table are found when using 1,2 and 3 dice. Why is p(x) decreasing as x increases?

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