2. A pub offers a '10 oz Steak Special'. If the steaks they use for these meals have uncooked weights that are normally distributed with a mean of 9.8 ounces and a standard deviation of 0.5 ounces, what is the probability that a customer will get:
I) A steak that has an uncooked weight of more than 10 ounces?
II) A steak that has an uncooked weight of more than 9.5 ounces?
III) A steak that has an uncooked weight of less than 10.5 ounces?
IV) In the population of uncooked steaks, what is:
a) The minimum weight of the heaviest 20% of the steaks?
b) The maximum weight of the lightest 10% of the steaks?
V) A group of four friends visit the pub and each of them orders a 'Steak Special'. What is the probability that the mean uncooked weight of the steaks they order is more than 10 ounces?
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.
For V, use Z = (score-mean)/SEm
SEm = SD/√n
To solve these questions, we will be using the properties of the normal distribution. The normal distribution is characterized by its mean (μ) and standard deviation (σ). In this case, the mean is 9.8 ounces and the standard deviation is 0.5 ounces.
I) To find the probability of getting a steak with an uncooked weight of more than 10 ounces, we need to calculate the area under the curve to the right of 10 ounces. Since the weights follow a normal distribution, we can use the z-score formula.
First, calculate the z-score:
z = (x - μ) / σ
z = (10 - 9.8) / 0.5
z = 0.4 / 0.5
z = 0.8
Using a z-table, we can find the probability associated with a z-score of 0.8. The table tells us that the area to the left of 0.8 is approximately 0.7881. Therefore, the probability of getting a steak with an uncooked weight of more than 10 ounces is 1 - 0.7881 = 0.2119, or 21.19%.
II) Similarly, for a steak with an uncooked weight of more than 9.5 ounces, we need to calculate the z-score:
z = (9.5 - 9.8) / 0.5
z = -0.3 / 0.5
z = -0.6
Using the z-table, we find that the area to the left of -0.6 is approximately 0.2743. Therefore, the probability of getting a steak with an uncooked weight of more than 9.5 ounces is 1 - 0.2743 = 0.7257, or 72.57%.
III) To find the probability of getting a steak with an uncooked weight of less than 10.5 ounces, we calculate the z-score:
z = (10.5 - 9.8) / 0.5
z = 0.7 / 0.5
z = 1.4
Using the z-table, we find that the area to the left of 1.4 is approximately 0.9192. Therefore, the probability of getting a steak with an uncooked weight of less than 10.5 ounces is 0.9192 or 91.92%.
IV)
a) To find the minimum weight of the heaviest 20% of the steaks, we need to calculate the z-score associated with the 80th percentile. Using the inverse z-score function on the z-table, we find that the z-score corresponding to the 80th percentile is approximately 0.8416.
Now, we can calculate the corresponding weight:
z = (x - μ) / σ
0.8416 = (x - 9.8) / 0.5
Solving for x:
x - 9.8 = 0.8416 * 0.5
x - 9.8 = 0.4208
x = 10.2208
Therefore, the minimum weight of the heaviest 20% of the steaks is approximately 10.2208 ounces.
b) Similarly, to find the maximum weight of the lightest 10% of the steaks, we need to calculate the z-score associated with the 10th percentile. Using the inverse z-score function on the z-table, we find that the z-score corresponding to the 10th percentile is approximately -1.2816.
Now, we can calculate the corresponding weight:
z = (x - μ) / σ
-1.2816 = (x - 9.8) / 0.5
Solving for x:
x - 9.8 = -1.2816 * 0.5
x - 9.8 = -0.6408
x = 9.1592
Therefore, the maximum weight of the lightest 10% of the steaks is approximately 9.1592 ounces.
V) To find the probability that the mean uncooked weight of the four steaks ordered is more than 10 ounces, we need to use the sampling distribution of the mean. The mean of the sampling distribution is equal to the population mean, which in this case is 9.8 ounces. The standard deviation of the sampling distribution is equal to the population standard deviation divided by the square root of the sample size.
σ(sample mean) = σ / √n
σ(sample mean) = 0.5 / √4
σ(sample mean) = 0.5 / 2
σ(sample mean) = 0.25
Now, we need to calculate the z-score for a mean of 10 ounces:
z = (x - μ) / σ
z = (10 - 9.8) / 0.25
z = 0.2 / 0.25
z = 0.8
Using the z-table, we find that the area to the left of 0.8 is approximately 0.7881. Therefore, the probability that the mean uncooked weight of the four steaks ordered is more than 10 ounces is 1 - 0.7881 = 0.2119, or 21.19%.
To solve these problems, we can use the standard normal distribution table or Excel functions to find the probabilities associated with the z-scores.
I) To find the probability that a customer will get a steak with uncooked weight more than 10 ounces, we need to find the area under the normal distribution curve to the right of 10 ounces. First, we need to standardize the value of 10 ounces into a z-score using the formula:
z = (x - μ) / σ
where x is the value we are interested in, μ is the mean, and σ is the standard deviation.
In this case, x = 10 ounces, μ = 9.8 ounces, and σ = 0.5 ounces. Substituting these values into the formula:
z = (10 - 9.8) / 0.5 = 0.4 / 0.5 = 0.8
Next, we can look up the corresponding area in the standard normal distribution table or use an Excel function like NORM.S.DIST(z,TRUE) to find the probability associated with a z-score of 0.8. Subtracting this probability from 1 will give us the probability that a customer will get a steak with uncooked weight more than 10 ounces.
II) To find the probability that a customer will get a steak with uncooked weight more than 9.5 ounces, we can follow a similar process. First, calculate the z-score:
z = (9.5 - 9.8) / 0.5 = -0.3 / 0.5 = -0.6
Look up the corresponding area or use an Excel function like NORM.S.DIST(z,TRUE) to find the probability associated with a z-score of -0.6. Subtracting this probability from 1 will give us the probability that a customer will get a steak with uncooked weight more than 9.5 ounces.
III) To find the probability that a customer will get a steak with uncooked weight less than 10.5 ounces, we can use the same method. Calculate the z-score:
z = (10.5 - 9.8) / 0.5 = 0.7 / 0.5 = 1.4
Look up the corresponding area or use an Excel function like NORM.S.DIST(z,TRUE) to find the probability associated with a z-score of 1.4. This will give us the probability that a customer will get a steak with uncooked weight less than 10.5 ounces.
IV)
a) To find the minimum weight of the heaviest 20% of the steaks, we need to find the z-score that corresponds to the 80th percentile (1 - 0.2 = 0.8). Using the standard normal distribution table or Excel function like NORM.S.INV(0.8), we can find the z-score associated with the 80th percentile. Finally, we can convert the z-score back to the actual weight using the formula:
x = (z * σ) + μ
b) To find the maximum weight of the lightest 10% of the steaks, we need to find the z-score that corresponds to the 10th percentile. Using the standard normal distribution table or Excel function like NORM.S.INV(0.1), we can find the z-score associated with the 10th percentile. Finally, we can convert the z-score back to the actual weight using the formula mentioned earlier.
V) To find the probability that the mean uncooked weight of the steaks the four friends order is more than 10 ounces, we can use the central limit theorem. The theorem states that the distribution of sample means will approach a normal distribution with the same mean as the population and a standard deviation equal to the population standard deviation divided by the square root of the sample size. In this case, the sample size is 4.
So, to find the probability, we will calculate the z-score using the formula:
z = (x̄ - μ) / (σ / sqrt(n))
where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, x̄ = 10 ounces, μ = 9.8 ounces, σ = 0.5 ounces, and n = 4. Substituting these values into the formula, we can calculate the z-score. Finally, we can look up the corresponding area or use an Excel function like NORM.S.DIST(z,TRUE) to find the probability associated with the z-score.