Asked by Michael
The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.52 inches and a standard deviation of 0.46 inches
(A) What percentage of the grapefruits in this orchard is larger than 5.45 inches?
(B) A random sample of 100 grapefruits is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 5.45 inches
(A) What percentage of the grapefruits in this orchard is larger than 5.45 inches?
(B) A random sample of 100 grapefruits is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 5.45 inches
Answers
Answered by
MathMate
Normalize!
Calculate Z=(X-μ)/σ
where
Z = Z score found in standard normal distribution tables
X = value of random variable, the diameter of 5.45 inchs
μ = mean diameter of population, 5.52 inches
σ = standard deviation of population, 0.46 inch.
Look up a standard normal distribution, which is usually given in percentage of population less than the Z-score, i.e. left tail (but check if this is the case).
What you need is the right tail, i.e. percentage of population greater than the Z-score. To get this from the table, subtract the left tail from ONE to get the right-tail.
Calculate Z=(X-μ)/σ
where
Z = Z score found in standard normal distribution tables
X = value of random variable, the diameter of 5.45 inchs
μ = mean diameter of population, 5.52 inches
σ = standard deviation of population, 0.46 inch.
Look up a standard normal distribution, which is usually given in percentage of population less than the Z-score, i.e. left tail (but check if this is the case).
What you need is the right tail, i.e. percentage of population greater than the Z-score. To get this from the table, subtract the left tail from ONE to get the right-tail.