PVC pipes is manufactured with a mean diameter of 1.01 inch and a standard deviation of 0.003 inch. Find the probability that a random sample of n = 9 sections of pipe will have a sample mean diameter greater than 1.009 inch and less than 1.012 inch. With solution please, Thank you.
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1.) PVC pipes is manufactured with a mean diameter of 1.01 inch and a standard deviation of 0.003 inch. Find the probability that a random sample of n = 9 sections of pipe will have a sample mean diameter greater than 1.009 inch and less than 1.012 inch.
With solution please, Thank you.
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- Assume that females have pulse rates that are normally distributed with a mean of u=74.0 beats per minute and a standard deviation of o = 12.5 beats per minute. if 4 adult females are randomly, find the probability that they have pulse rates with a mean between 67 beats per minute and 81 beats per minute.The thickness of a part is to have an upper specification of 0.925 and a lower specification of 0.870 mm. The average of the process is currently 0.917 with a standard deviation of 0.005. Determine the percentage of product above 0.90 mm.Answer in 30minutes
- Assume that adults have IQ scores that are normally distributed with a mean of 99.9 and a standard deviation of 24.5. find the probability that a randomly selected adult has an IQ greater than 134.1. (round to four decimal places as needed)According to a poll, a shopper spends an average of 45 minutes at the supermarket for each trip, with a standard deviation of 12 minutes. The variable x represents the duration of time spent in the shop, which is normally distributed. A customer walks inside the business. Determine the likelihood that the consumer will spend between 25 and 55 minutes in the store.Tall Dutch Women: In contrast, the mean adult female height in the Netherlands is 169 cm with a standard deviation of 7.1 cm and also follows a normal distribution. Using the empirical rule, (showing your calculations and not using the online calculator,) calculate what percentage of the Dutch adult female population would be between 154.8 cm and 161.9 cm tall. Express your answer as a percentage rounded off to 2 decimal places.
- A company that makes cola drinks states that the mean caffeine content per 12-ounce bottle of cola is 55 milligrams. You want to test this claim. During your tests, you find that a random sample of thirty 12-ounce bottles cola has a mean caffeine content of 57.4 milligrams. Assume the population is normally distributed and the population standard deviation is 6.9 milligrams. At a = 0.04, can you reject the company's claim? Complete parts (a) through (e). C (a) Identify Ho and H₂. Choose the correct answer below. OA. Ho: 257.4 OB. Ho: H=57.4 H_:μ#57.4 H₂H<57.4 C. Ho: H=55 H₂:μ#55 OD. Ho: μ255 H:H<55 OE. Ho: 57.4 OF. Ho: *55 H₂:μ=57.4 Ημ= 55 (b) Find the critical value(s). Select the correct choice below and fill in the answer box within your choice. (Round to two decimal places as needed.) A. The critical values are ± 2.05. OB. The critical value is. Identify the rejection region(s). Choose the correct answer below. OA OB. O C. O Fail to reject Ho Fail to reject Ho Fail to reject Ho Q X…Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find the probability of obtaining a reading between 0°C and 2.679°C.The average price of a kilogram of ham is P1,300. Assume that the standard deviation is P150. If a random sample of 30 one-kilogram packages is selected, find the probability that the mean sample will be less than P1,160. Show the complete solution.
- A company that makes cola drinks states that the mean caffeine content per 12-ounce bottle of cola is 55 milligrams. You want to test this claim. During your tests, you find that a random sample of thirty 12-ounce bottles of cola has a mean caffeine content of 57.4 milligrams. Assume the population is normally distributed and the population standard deviation is 6.9 milligrams. At a=0.04, can you reject the company's claim? Complete parts (a) through (e). (a) Identify Ho and H₂. Choose the correct answer below. OA. Ho: 257.4 H₂:μ< 57.4 OB. Ho: H=57.4 H₂:μ*57.4 C. Ho: H=55 H₂:55 OD. Ho: 255 H₂H<55 OE. Ho: 57.4 Η μ = 57.4 OF. Ho: *55 H₂: H=55 (b) Find the critical value(s). Select the correct choice below and fill in the answer box within your choice. (Round to two decimal places as needed.) A. The critical values are ± OB. The critical value isSuppose the masses of a batch of 2000 potatoes have a mean of 60 grams with a standard deviation of 8 grams. Assuming the weights have a normal distribution, predict the number of potatoes that would have masses between 45 and 55 grams. Sketch the distribution curve, shade the required probability, and show horizontal scales the population parameters and the z-scores. Solve the problemThe weights of cans of Ocean brand tuna are supposed to have a net weight of 6 ounces. The manufacturer tells you that the net weight is actually a Normal random variable with a mean of 5.95 ounces and a standard deviation of 0.2 ounces. Suppose that you draw a random sample of 42 cans. Parti) Suppose the number of cans drawn is doubled. How will the standard deviation sample mean weight change? A. It will decrease by a factor of √2. B. It will increase by a factor of √2. C. It will increase by a factor of 2. D. It will decrease by a factor of 2. E. It will remain unchanged. Part ii) Suppose the number of cans drawn is doubled. How will the mean of the sample mean weight change? ▸ A. It will increase by a factor of √2. B. It will decrease by a factor of 2. C. It will increase by a factor of 2. D. It will decrease by a factor of √2. E. It will remain unchanged. Part iii) Consider the statement: The distribution of the mean weight of the sampled cans of Ocean brand tuna is Normal." A. It…
