A study shows that 30 percent of children in a given developing country are malnourished. Using normal approximation, find the probabilities that in a random sample of 25 children from this region, the number of malnourished children is a. higher than 5; b. between 4 and 8.
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- A population has a mean of 50, and the standard deviation is 10. You take several samples and calculate sample means. 95% of these sample means lie between 46 and 54. What was the size of each sample? n= 1) 5 2) 4 3) 10 4) 25have 15mins help ASAP!Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 200 eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24%, 22% of them voted. The probability that fewer than 49 voted.
- Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 157 eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted. Probability that fewer than 41 votedAccording to a candy company, packages of a certain candy contain 24% orange candies. Find the approximate probability that the random sample of 300 candies will contain 29% or more orange candies. Using a normal approximation, what is the probability that at least 29% of 300 randomly sampled candies will be orange?Hershey Kiss Weight in Oz. 1 Jason, a Quality Process Engineer for Hershey Chocolates, pulled 16 random kisses from production and recorded their weights as shown. Jason uses the production number for each kiss in production to ensure that all of the kisses in production have an equal chance of being selected in his random number generator. .1618 .1607 .1613 A: Calculate a point estimate, x, for the population mean, u. .1607 .1610 1611 1609 .1597 .1586 .1604 B: Construct and interpret a 95% CI (Confidence Interval) for u. .1607 (Hint: Use s from your calculator's descriptive statistics.) .1606 1611 .1620 .1610 .1609 C: Construct and interpret a 99% CI (Confidence Interval) for µ.
- In a survey of 175 females who recently completed high school, 76% were enrolled in college. In a survey of 150 males who recently completed high school, 72% were enrolled in college. At a = 0.05, can you reject the claim that there is no difference in the proportion of college enrollees between the two groups? Assume the random samples are independent. Complete parts (a) through (e). O A. Ho: P1 #P2 Ha: P1 = P2 O B. Ho: P1 P2 OF. Ho: P1 = P2 O D. Ho: P1 > P2 Ha: P1 SP2 Ha: P1 O B. z 1.96 O D.use a normal approximation to find the probability of the indicated number of voters. in the case, assume that 136 eligible voters aged 18-24 are randomly selected. suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted. probability that exactly 36 voted.Nationally, about 11% of the total U.S. wheat crop is destroyed each year by hail.† An insurance company is studying wheat hail damage claims in a county in Colorado. A random sample of 16 claims in the county reported the percentage of their wheat lost to hail. 16 6 8 9 10 18 14 13 9 10 25 18 13 8 14 3 The sample mean is x = 12.1%. Let x be a random variable that represents the percentage of wheat crop in that county lost to hail. Assume that x has a normal distribution and σ = 5.0%. Do these data indicate that the percentage of wheat crop lost to hail in that county is different (either way) from the national mean of 11%? Use ? = 0.01. (a) What is the level of significance? Compute the z value of the sample test statistic. (Round your answer to two decimal places.)(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the a = 0.05 level of significance. Normal probability plots indicate that the data are approximately normal and boxplots indicate that there are no outliers. State the null and alternative hypotheses. Let population 1 be carpeted rooms and population 2 be uncarpeted rooms. A. Ho: H1= H2 H1: H1> H2 B. Ho: H1 H2 c. Hoi H1= H2 H1: H1The proportion of items in a population that possess a specific attribute is known to be 0.25 a. If a simple random sample of size n=100 is selected and the proportion of items in the sample that contain the attribute of interest is 0.30, what is the sampling error b. Referring to part a, what is the probability that a sample size of 100 would have a sample proportion of 0.30 or less if the population proportion is 0.25Nationally, about 11% of the total U.S. wheat crop is destroyed each year by hail.† An insurance company is studying wheat hail damage claims in a county in Colorado. A random sample of 16 claims in the county reported the percentage of their wheat lost to hail. 15 7 7 13 13 20 13 12 5 10 26 22 13 7 11 3 The sample mean is x = 12.3%. Let x be a random variable that represents the percentage of wheat crop in that county lost to hail. Assume that x has a normal distribution and σ = 5.0%. Do these data indicate that the percentage of wheat crop lost to hail in that county is different (either way) from the national mean of 11%? Use α = 0.01. (a) What is the level of significance? Compute the z value of the sample test statistic. (Round your answer to two decimal places.)(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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