Given a normally distributed variable x with mean 18 and standard deviation 2.5, find: P (x < 15) the value of x such that P (x < x) = 0.2236 the value of x such that P (x > x) = 0.1814 P (17 < x < 21)
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- A company claims that the number of defective items manufactured during each run of making 100 of their products is independent of the number from other runs and that the proportion of defectives is not more than 3%. Assuming that the defective rate for each run is 3% Which of the following can be used to determine whether x = 8 is unusually a high number of defective items on the next run of 100 of their products?Let x = red blood cell (RBC) count in millions per cubic millimeter of whole blood. For healthy females, x has an approximately normal distribution with mean ? = 4.3 and standard deviation ? = 0.5. (a) Convert the x interval, 4.5 < x, to a z interval. (Round your answer to two decimal places.) < z(b) Convert the x interval, x < 4.2, to a z interval. (Round your answer to two decimal places.)z < (c) Convert the x interval, 4.0 < x < 5.5, to a z interval. (Round your answers to two decimal places.) < z < (d) Convert the z interval, z < −1.44, to an x interval. (Round your answer to one decimal place.)x < (e) Convert the z interval, 1.28 < z, to an x interval. (Round your answer to one decimal place.) < x(f) Convert the z interval, −2.25 < z < −1.00, to an x interval. (Round your answers to one decimal place.) < x <6.9 To enter the cumulative normal distribution table Þ(Z), calculate Z and Z at the spec limits. upper lower Þ(2) = 0.9772 and Þ(-2) = 0.0228, the area to the right of the USL plus to the left of the LSL is 0.0456.
- Assume x is normally distributed with a mean of 2 and a standard deviation of 2. Find: a.P(X=2) b.P(X<2) c.P(X>=2) d.P(2 < X < 4)a b Absolute standard deviation = Coefficient of variation = y = 1.20 (+0.02) x 10-8-3.60(±0.2) × 10-⁹ x Result = Absolute standard deviation = Coefficient of variation = C Result = d y=90.95 (±0.08) - 89.40(±0.06) +0.200(+0.004) Absolute standard deviation == Coefficient of variation = Result = ± y= y = 0.0040 (+0.0005) x 10.28 (+0.02) × 395(+1) Result = ± Absolute standard deviation = Coefficient of variation = % 329(+0.03) x 10-14 1.47 (+0.04) x 10-16 % % H % ±Suppose the mean value for Rh is 1,106, the mean value for X,R, is 2,200, the standard deviation for Rh is 427.1 and the standard deviation for X,R, is 325.3. What then is the value for the standard deviation of the limit state function V (answer to 1 decimal place)?
- If x is normally distributed with population mean=20.0 and population standard deviation = 4.0, determine P(16.0 less than/equal to x less than/equal to 24.0)Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean μ = 27.0 kilograms and standard deviation σ = 3.1 kilograms. Let x be the weight of a fawn in kilograms. For parts (a), (b), and (c), convert the x intervals to z intervals. (For each answer, enter a number. Round your answers to two decimal places.) (a) x < 30z < (b) 19 < x (Fill in the blank. A blank is represented by _____.)_____ < z (c) 32 < x < 35 (Fill in the blanks. A blank is represented by _____. There are two answer blanks.)_____ < z < _____first blank second blank For parts (d), (e), and (f), convert the z intervals to x intervals. (For each answer, enter a number. Round your answers to one decimal place.) (d) −2.17 < z (Fill in the blank. A blank is represented by _____.)_____ < x (e) z < 1.28x < (f) −1.99 < z < 1.44 (Fill in the blanks. A blank is represented by _____. There are two answer…Given the normally distributed variable X with mean 18 and standard deviation 2.5, find (a) P(X k) = 0.1814; (d) P(15SEE MORE QUESTIONS
