Let z be a normal random variable with a mean of 0 and a standard deviation of 1. Determine P(z)<=(1.4)
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Let z be a normal random variable with a mean of 0 and a standard deviation of 1. Determine P(z)<=(1.4)
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- PLEASE HELP ME WITH THIS. THANK YOU!Use the formula to find the standard error of the distribution of differences in sample means, x⎯⎯1−x⎯⎯2x¯1-x¯2.Samples of size 100 from Population 1 with mean 95 and standard deviation 10 and samples of size 80 from Population 2 with mean 76 and standard deviation 17Round your answer for the standard error to two decimal places.standard error =Pyramid Lake is on the Palute Indian Reservation in Nevada. The lake is famous for cutthroat trout. Suppose a tells you that the average length of trout caught in Pyramid Lake Is mu = 19 Inches. Howevera survey reported that of a random sample of 46 fish caught, the mean length was overline x =18.6 lnch , with estimated standard s = 2.7 inches. these data Indicate that the average length of a trout caught in Pyramid Lake less than mu = 19 inches? Use a=0.05 Please answer ONLY parts D & E
- asap pleaseLet x be a random variable that represents red blood cell count (RBC) in millions of cells per cubic millimeter of whole blood. Then x has a distribution that is approximately normal. For the population of healthy female adults, suppose the mean of the x distribution is about 4.78. Suppose that a female patient has taken six laboratory blood tests over the past several months and that the RBC count data sent to the patient's doctor are as follows. 4.9 4.2 4.5 4.1 4.4 4.3 (i) Use a calculator with sample mean and standard deviation keys to find x and s. (Round your answers to two decimal places.) x = s = (ii) Do the given data indicate that the population mean RBC count for this patient is lower than 4.78? Use ? = 0.05. (a) What is the level of significance?State the null and alternate hypotheses. H0: ? < 4.78; H1: ? = 4.78H0: ? = 4.78; H1: ? ≠ 4.78 H0: ? = 4.78; H1: ? < 4.78H0: ? = 4.78; H1: ? > 4.78H0: ? > 4.78; H1: ? = 4.78 (b) What sampling…A standardized exam's scores are normally distributed. In a recent year, the mean test score was 1463 and the standard deviation was 314. The test scores of four students selected at random are 1870, 1220, 2190, and 1340. Find the z-scores ← that correspond to each value and determine whether any of the values are unusual. F1 The z-score for 1870 is (Round to two decimal places as needed.) The z-score for 1220 is. (Round to two decimal places as needed.) The z-score for 2190 is (Round to two decimal places as needed.) The z-score for 1340 is. (Round to two decimal places as needed.) Which values, if any, are unusual? Select the correct choice below and, if necessary, fill in the answer box within your choice. OA. The unusual value(s) is/are. (Use a comma to separate answers as needed.) OB. None of the values are unusual. F2 80 F3 000 000 F4 F5 ^ MacBook Air F6 & A D F7 * DII F8 DD F9 ) J 운 F10 I a F11 + Next F12 delete
- The time spent waiting in the line is approximately normally distributed. The mean waiting time is 6 minutes and the variance of the waiting time is 1. Find the probability that a person will walt for less than 7 minutes. Round your answer to four decimal places. Answer esc If you would like to look up the value in a table, select the table you want to view, then either click the cell at the intersection of the row and column or use the arrow keys to find the appropriate cell in the table and select it using the Space key. 1 hp angle. q a Z F @ 2 W S 320 3 e d C $ 4 r f % 5 V t g Normal Table - to-z Normal Table - to z 6 O Oll b y hp h & 7 O n u J - * 8 N O i m 9 k * O : Tables ctrl Keypa Keyboard Shortc Submit Answer { D + = ? Oct 27 11 } 5:21 8 back 1 USE YOUR SMARTIN Reviews - Videog Spass SunThe 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…Help fast please
- please answer this question within 30 minutes. i will upvote.***Only do this question if using Minitab and show images*** Let X1, X2, X3, . . . , X200 denote the weights of 200 independent and identically distributed bags of candy corn. If the mean weight of each bag is 2 lb. and its standard deviation is 0.07 lb., determine the probability that the average of 200 bags weighs between 1.997 lb. and 2.06 lb. Show your answer using Minitab.+ I/ * 00 A company manufactures light bulbs. The company wants the bulbs to have a mean life span of 997 hours. This average is maintained by periodically testing random samples of 16 light bulbs. If the t-value falls between -th an and t, go then the company will be satisfied that it is manufacturing acceptable light bulbs. For a random sample, the mean life span of the sample is 1002 hours and the standard deviation is 29 hours. Assume that life spans are approximately normally distributed. Is the company making acceptable light bulbs? Explain. and to 90 The company making acceptable light bulbs because the t-value for the sample is t= (Round to two decimal places as needed.) Get more help- Clear all Check answer 3:44 PM 64°F Sunny A 11/23/2021 PrtSc F5. F10 F11 F12 4. 5. 9. 6. K. B. M.
