Sorry the full question is A loaf of bread is normally distributed with a mean of22 oz. and a standard deviation of 0.5 oz.
a, what is the probability that a loaf is btwn 22.75and 23.00?
B, 5% of the loafs will weigh less than_____ ounces?
C, assuming 200 loaves are baked on a given day, how many of these will weigh more than 23 ounces? thanks for the help...
a. Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
b. Use same table to get Z related to .05 in smaller portion and insert values in above equation to get score.
c. Use table to find proportion and multiply by 200.
a) To find the probability that a loaf of bread is between 22.75 and 23.00 ounces, we need to calculate the area under the normal distribution curve between these two values.
First, we need to standardize the values by subtracting the mean and dividing by the standard deviation:
Standardized value for 22.75 ounces:
Z1 = (22.75 - 22) / 0.5 = 1.5
Standardized value for 23.00 ounces:
Z2 = (23.00 - 22) / 0.5 = 2
Next, we use a standard normal distribution table or a calculator to find the probability associated with each standardized value. Subtract the smaller probability from the larger probability to find the probability between the two values:
P(22.75 < X < 23.00) = P(1.5 < Z < 2)
Using a standard normal distribution table, we find the corresponding probabilities:
P(Z < 1.5) ≈ 0.9332
P(Z < 2) ≈ 0.9772
So the probability of a loaf of bread weighing between 22.75 and 23.00 ounces is approximately 0.9772 - 0.9332 ≈ 0.044.
b) To find the weight at which 5% of the loaves will weigh less than, we need to find the corresponding Z-score.
Using a standard normal distribution table or calculator, we find the Z-score associated with a cumulative probability of 5%. This Z-score represents the number of standard deviations below the mean.
Z = -1.645 (approximately)
Now we can use the Z-score formula to find the corresponding weight:
Z = (X - Mean) / Standard Deviation
Substituting the known values:
-1.645 = (X - 22) / 0.5
Solving for X, we have:
X - 22 = -1.645 * 0.5
X - 22 = -0.8225
X ≈ 22 - 0.8225 ≈ 21.1775
So 5% of the loaves will weigh less than approximately 21.18 ounces.
c) To find the number of loaves that weigh more than 23 ounces out of 200 loaves, we can use the cumulative probability associated with a Z-score of 23:
Z = (23 - 22) / 0.5 = 2
Using a standard normal distribution table or calculator, we find the probability associated with a Z-score of 2:
P(Z > 2) ≈ 1 - P(Z < 2) ≈ 1 - 0.9772 ≈ 0.0228
This means that approximately 0.0228 or 2.28% of the loaves will weigh more than 23 ounces.
To find the number of loaves that satisfy this condition, we multiply the percentage by the total number of loaves:
Number of loaves weighing more than 23 ounces = 0.0228 * 200 ≈ 4.56
So, approximately 4 or 5 loaves will weigh more than 23 ounces out of 200 loaves.