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A population of values has a normal distribution with u = 12 and o = 90.4. If a random sample of size

n = 14 is selected,
a. Find the probability that a single randomly selected value is greater than -24.2. Round your answer to four decimals.
P(X > -24.2) =
b. Find the probability that a sample of size n = 14 is randomly selected with a mean greater than
-24.2. Round your answer to four decimals.
P(M > -24.2) =

User Cade Roux
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2 Answers

1 vote
it’s -25.3 if I’m right but
User Parthapratim Neog
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6 votes

Final answer:

To find the probability that a single randomly selected value is greater than -24.2, standardize the value and use the standard normal distribution table. The probability is approximately 0.3553. To find the probability that a sample of size n = 14 has a mean greater than -24.2, calculate the standard error of the mean and standardize the value. The probability is approximately 0.9302.

Step-by-step explanation:

To find the probability, we need to standardize the value -24.2 by subtracting the mean (u) and dividing by the standard deviation (o). The standardized value is z = (-24.2 - 12) / 90.4 = -0.3683. Now we can use the standard normal distribution table or a calculator to find the probability. From the table, the probability of a randomly selected value being greater than -0.3683 is approximately 0.3553.

To find the probability that a sample of size n = 14 has a mean greater than -24.2, we need to calculate the standard error of the mean (SE) by dividing the standard deviation (o) by the square root of the sample size (n). The SE is calculated as 90.4 / sqrt(14) = 24.1952. Now we can standardize the value -24.2 by subtracting the mean (u) and dividing by the SE. The standardized value is z = (-24.2 - 12) / 24.1952 = -1.4817. Using the standard normal distribution table or a calculator, the probability of a sample mean being greater than -1.4817 is approximately 0.9302.

User Corbell
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