Totally confused on how to solve this problem. It readsas: Suppose x is a normal

Question

Totally confused on how to solve this problem. It readsas: Suppose x is a normally distriuted random variable with = 30 and = 8. Find a value of the randomvariable, call it x0, such that a. P(x > x0) = .5 b. P(x < x0) = .025 c. 10 % of the values of x are less thanx0 d. 80 % of the values of x are less than x0 Thank you!           Totally confused on how to solve this problem. It readsas: Suppose x is a normally distriuted random variable with = 30 and = 8. Find a value of the randomvariable, call it x0, such that a. P(x > x0) = .5 b. P(x < x0) = .025 c. 10 % of the values of x are less thanx0 d. 80 % of the values of x are less than x0 Thank you! Thank you!          

Explanation / Answer

a)   The value would be x = 30 since z =(30-30)/8 = 0   and P(z 0) = 0.5 b) Using the table backwards, we find a p-value close to0.025. The z-score that corresponds to this p-value is-1.96. So, (x -30)/8 = -1.96  ----->    x = 14.32 c) Using the table backwards, we find a p-value close to0.1000. The z-score that corresponds to this p-value is-1.28. So, (x -30)/8 = -1.28  ----->    x = 19.76

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