Suppose a random variable has the normal distribution such that X(82,4.8)X(82,4.

Question

Suppose a random variable has the normal distribution such that X(82,4.8)X(82,4.8). Compute the probability that the random variable will take on a value less than 89.2. (Include a leading zero and 4 decimal places.)

Suppose a random variable has the normal distribution such that X(82,4.8)X(82,4.8). Compute the probability that the random variable will take on a value greater than 78.4. (Include a leading zero and 4 decimal places.)

Suppose a random variable has the normal distribution such that X(82,4.8)X(82,4.8). Compute the probability that the random variable will take on a value of at least 78.4. (Include a leading zero and 4 decimal places.)

Suppose a random variable has the normal distribution such that X(82,4.8)X(82,4.8). Compute the probability that the random variable will take on a value between 83.2 and 88.0. (Include a leading zero and 4 decimal places.)

Suppose a random variable has the normal distribution such that X(82,4.8)X(82,4.8). Compute the probability that the random variable will take on a value less than 80.0 or more than 90.0. (Include a leading zero and 4 decimal places.)

Suppose a random variable has the normal distribution such that X(82,4.8)X(82,4.8). Compute the probability that the random variable will be 1.5 standard deviations or more to the right. (Include a leading zero and 4 decimal places.)

Explanation / Answer

AS per chegg policies i am answering 4 parts

Suppose a random variable has the normal distribution such that X(82,4.8)X(82,4.8). Compute the probability that the random variable will take on a value less than 89.2. (Include a leading zero and 4 decimal places.) p(x<89.2) = p(z<(89.2-82)/4.8) =NORMDIST(89.2,82,4.8,TRUE) 0.9332 Suppose a random variable has the normal distribution such that X(82,4.8)X(82,4.8). Compute the probability that the random variable will take on a value greater than 78.4. (Include a leading zero and 4 decimal places.) P(X>78.4) = 1-P(X<78.4) 1-P(Z<(78.4-82)/4.8) =1-NORMDIST(78.4,82,4.8,TRUE) 0.7734 Suppose a random variable has the normal distribution such that X(82,4.8)X(82,4.8). Compute the probability that the random variable will take on a value of at least 78.4. (Include a leading zero and 4 decimal places.) P(X>=78.4) = 1-P(X<78.4) 1-P(Z<(78.4-82)/4.8) =1-NORMDIST(78.4,82,4.8,TRUE) 0.7734 Suppose a random variable has the normal distribution such that X(82,4.8)X(82,4.8). Compute the probability that the random variable will take on a value between 83.2 and 88.0. (Include a leading zero and 4 decimal places.) P(83.2<X<88) P(X<88)-P(X<83.2) P(Z<(88-82)/4.8)-P(Z<(83.2-82)/4.8) =NORMDIST(88,82,4.8,TRUE)-NORMDIST(83.2,82,4.8,TRUE) 0.2956

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