Question #6 Part C # Standard Normal Distribution TableThis document provides a comprehensive standard normal distribution table used to find probabilities related to the standard normal curve, a crucial concept in statistics.## DescriptionThe table is divided into two sections, each displaying areas under the normal curve for positive and negative z-scores. These areas represent the probability that a standard normal random variable is less than the given z-value. The table includes:- **Z-scores**: These values are represented in the leftmost column. Each entry is expressed with one decimal place.- **Probabilities**: The intersecting values within the table correspond to the area (probability) under the curve to the left of the given z-score. Each column header from 0.00 to 0.09 adds a hundredth to the z-score in the row.### Graphical ExplanationTwo small graphs above the tables illustrate the normal curve with shaded areas:- **Curve**: Represents the bell-shaped, standard normal distribution.- **Shaded Area**: Represents the cumulative probability from the z-score to the left, matching the tabular values. The graph visually demonstrates how the area under the curve corresponds to the values found in the table.## UsageTo find the probability associated with a specific z-score:1. Locate the z-score row (e.g., 0.1) and the column with the correct hundredths place (e.g., 0.03).2. Intersect the row and column to find the probability.3. Use the left table for positive z-scores and the right table for negative z-scores.The table is essential for statistical analysis, including hypothesis testing and confidence interval construction. **Text Transcription:**A random sample of 1017 adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1017 adults surveyed, 515 indicated that televisions are a luxury they could do without. Complete parts (a) through (e) below.- Click here to view the standard normal distribution table (page 1).- Click here to view the standard normal distribution table (page 2).**(b) Verify that the requirements for constructing a confidence interval about p are satisfied.**The sample is stated to be a simple random sample, the value of \( \hat{p}(1-\hat{p}) \) is 254.213, which is greater than or equal to 10, and the sample size can be assumed to be less than or equal to 5% of the population size. (Round to three decimal places as needed.)**(c) Construct and interpret a 95% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without. Select the correct choice below and fill in any answer boxes within your choice.**(Type integers or decimals rounded to three decimal places as needed. Use ascending order.)- \( \bigcirc \) A. There is a \(\_\_\_\_\_\_\) % probability the proportion of adults in the country who believe that televisions are a luxury they could do without is between \(\_\_\_\_\_\_\) and \(\_\_\_\_\_\_\).- \( \bigcirc \) B. We are \(\_\_\_\_\_\_\) % confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between \(\_\_\_\_\_\_\) and \(\_\_\_\_\_\_\).

Given Data Sample Size, n = 1017 Number of successes, x = 515.0 confidence level,c= 0.95solution : Significance level, α =1- 0.95 = 0.05 Point estimate p̂ = x/n = 515.0 / 1017 = 0.506391 Critical value critical value at α = 0.05 is Zα/2 = 1.96 ....from the standard normal distribution table 95% confidence interval = 0.506391 ± 1.96 x √ {( 0.506391 (1- 0.506391 )/ 1017 } = 0.506391 ± 0.030728 =( 0.506391 - 0.030728 , 0.506391 + 0.030728 ) =( 0.476 , 0.537 ) We are 95 % confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.476 and 0.537