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- 4. A medical statistician wants to estimate the average weight loss of people who are on a new diet plan. In a preliminary study, he guesses that the standard deviation of weight loss is about 10 pounds. How large a sample should he take to estimate the mean weight loss to within 2 pounds, with a 90% confidence.1. The amount of time that people spend at Grover Hot Springs is normally distributed with a mean of 76 minutes and a standard deviation of 16 minutes. Suppose one person at the hot springs is randomly chosen. Let X = the amount of time that person spent at Grover Hot Springs. Round all answers to 4 decimal places where possible.A. Find the(IQR) for time spent at the hot springs.Q1, Q3 IRQ: minutes 2. The average American man consumes 9.5 grams of sodium each day. Suppose that the sodium consumption of American men is normally distributed with a standard deviation of 1 grams. Suppose an American man is randomly chosen. Let X = the amount of sodium consumed. Round all numeric answers to 4 decimal places where possible.B. The middle 20% of American men consume between what two weights of sodium? Low___High____The daily electric consumption of two houses were recorded. House A has consumes 375W (mean) the standard deviation is 15W. While House B uses 250W (mean), the standard deviation is 22W. 1. Find the probability for each house that the electric usage will not exceed 400W.2. Find the probability that both houses will exceed 400W in usage.3.Given the probability of 0.02, how much electricity must House B consume to avoid exceeding?
- 1. "The weights of 1,000 children, in average, is 45kg with standard deviation of 18kg. Suppose the weights are normally distributed, how many children weigh between 53kg and 58kg?"a. 670b. 236c. 94d. 7646. A distribution with a µ = 55 and o = 6 is standardized so that the new mean and standard deviation will be u = 50 and o = 10. When the new distribution is standardized, what value will be obtained for a score of x = 52 from the original distribution? a. X= 45 b. X = 47 C. X= 52 d. X = 533. The average height of a male student in the EDA class of Engr. Ventura in PLM has historically been 1.625 meters with a standard deviation of 0.07 meter. Is there a reason to believe that there has been a change in the average height of male EDA students under Engr. Ventura if a random sample of 50 males in the present EDA class has an average height of 1.652 meters? Assume the standard deviation remains the same.
- 5. The mean birth weight for babies born one month early is 2630 grams. Assume that the population has a standard deviation of 220 grams. Sketch the distribution of birth weights (in grams) of children born one month early. Calculate probability of a child born one month early with a birth weight between 2100 grams and 2900 grams. Find the weight which corresponds to the top 25% of birth weights for babies born one month early.5.6.
- Suppose the mean blood pressure for people in a certain country is 130 mmHg with a standard deviation of 24 mmHg. Blood pressure is normally distributed. State the random variable. The mean blood pressure of people in the country. The standard deviation of blood pressures of people in the country. O The blood pressure of a person in the country. Suppose a sample of size 10 is taken. State the shape of the distribution of the sample mean. The shape of the sampling distribution of the sample mean is unknown since the population of the random variable is normally distributed and the sample size is less than 30. The sampling distribution of the sample mean is normally distributed since the population is normally distributed. The sampling distribution of the sample mean is unknown since the sample size is less than 30. Suppose a sample of size 10 is taken. State the mean of the sample mean. Suppose a sample of size 10 is taken. State the standard deviation of the sample mean. Round to two…4. Jim wishes to estimate the average number of days students are absent from his STA 205 in a semester. In a random sample of 36 of her STA 205 students the mean number of absences was 2.853 days with a standard deviation of 1.878 days. Find the interval that is three standard deviations from the mean. Interpret this interval using either the Empirical Rule or Chebyshev's Rule (whichever is most appropriate for this distribution). Estimate the mean number of days students are absent from Jim's STA 205 using 90% confidence. Two-tail probability One-tail probability 0.20 0.10 0.05 0.02 0.01 0.10 0.05 0.025 0.01 0.005 Table T df df 30 1.310 1.697 2.042 2.457 2.750 30 32 1.309 1.694 2.037 2.449 2.738 32 35 1.306 1.690 2.030 2.438 2.725 35 1.684 2.021 2.423 2.704 2.690 40 1.303 40 45 1.301 1.679 2.014 2.412 45