Score: 0 of 1 pt 8.4.28 A random sample of size 49 is taken from a normal popula

Question

Score: 0 of 1 pt 8.4.28 A random sample of size 49 is taken from a normal population having a mean of 90 and a standard deviation of 5. A second random sample of size 36 | 7 of 12 (9 complete) HYV Score: 59.72%, 7.17 of 12 pts is taken from a different normal population having a mean of 85 and a standard the 49 measurements will exceed the sample mean computed trom the 36 measurements by at least 3.3 but less than 6.2. Assume the difference of the means to be measured to the nearest tenth Click here to view page 1 of the standard Help deviation of 7. Find the probability that the sample mean computed from The probability is 0.7031 (Round to tour decimal places as needed.)

Explanation / Answer

here mean difference between sample =90-85 =5

and std error of difference =(52/49+72/36)1/2 =1.3680

( please try this andrevert)

probability = P(3.3<X<6.2) = P(-1.2427<Z<0.8772)= 0.8098-0.1070= 0.7028

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