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
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.
To find the answer to these questions, we will use the concept of the standard normal distribution, also known as the Z-distribution. The Z-distribution is a normal distribution with a mean of 0 and a standard deviation of 1.
(A) To find the percentage of grapefruits larger than 5.45 inches, we need to calculate the z-score for 5.45 using the given mean and standard deviation.
The formula to calculate the z-score is:
z = (x - μ) / σ
where:
- x is the value we're interested in (in this case, 5.45 inches)
- μ is the population mean (5.52 inches)
- σ is the standard deviation (0.46 inches)
Plugging in the values, we have:
z = (5.45 - 5.52) / 0.46
Calculating this expression, we find:
z ≈ -0.15
Next, we need to find the corresponding area under the standard normal distribution curve for this z-score. This can be done using a Z-table or a statistical calculator.
Looking up the z-score of -0.15 in a Z-table, we find that the area to the left of -0.15 is approximately 0.4382.
Since we want the percentage of grapefruits that are larger than 5.45 inches, we subtract this value from 1:
Percentage = 1 - 0.4382 ≈ 0.5618
So, approximately 56.18% of the grapefruits in the orchard are larger than 5.45 inches.
(B) To find the probability that the sample mean is greater than 5.45 inches, we need to use the Central Limit Theorem. According to the Central Limit Theorem, for a large enough sample size, the distribution of sample means approaches a normal distribution, regardless of the shape of the population distribution.
Given that the sample size is 100, we can assume that the distribution of the sample mean will be approximately normal.
The mean of the sample mean is the same as the population mean, which is 5.52 inches.
The standard deviation of the sample mean (also known as the Standard Error) can be calculated using the formula:
SE = σ / √n
where:
- σ is the population standard deviation (0.46 inches)
- n is the sample size (100)
Plugging in the values, we have:
SE = 0.46 / √100
Calculating this expression, we find:
SE = 0.046
Next, we need to calculate the z-score using the formula:
z = (x - μ) / SE
where:
- x is the value we're interested in (in this case, 5.45 inches)
- μ is the population mean (5.52 inches)
- SE is the standard error (0.046)
Plugging in the values, we have:
z = (5.45 - 5.52) / 0.046
Calculating this expression, we find:
z ≈ -1.52
Again, we need to find the corresponding area under the standard normal distribution curve for this z-score (-1.52).
Looking up the z-score of -1.52 in a Z-table, we find that the area to the left of -1.52 is approximately 0.0630.
Since we want the probability that the sample mean is greater than 5.45 inches, we subtract this value from 1:
Probability = 1 - 0.0630 ≈ 0.9370
So, there is approximately a 93.70% probability that the sample mean is greater than 5.45 inches.