9.)The actual weight of 2-pound sacks of salted peanuts is found to be normally distributed with a mean
equal to 2.04 pounds and a standard deviation of 0.25 pounds. Given this information, the probability of a
sack weighing more than 2.40 pounds is 0.4251. **TRUE OR FALSE??**
13.)A university computer lab manager wishes to estimate the mean time that students stay in the lab per
visit. She believes that the population standard deviation would be no larger than 10 minutes. Further, she
wishes to have a confidence level of 90 percent and a margin of error of ± 2.00 minutes. Given this, the
sample size that she uses is approximately 60 students.
**TRUE OR FALSE?**
Check 9) using z-scores:
z = (x - mean)/sd
z = (2.40 - 2.04)/0.25
z = 1.44
Using a z-table, probability = 0.0749
Answer: False
Check 13) using margin of error:
Margin of error = z-value * (sd/√n)
2.00 = 1.645 (10/√n)
n is approximately 68 (rounded up)
Answer: Maybe true (depends on how close you want to approximate).
9.) TRUE
To find the probability of a sack weighing more than 2.40 pounds, we need to calculate the z-score and find the corresponding probability from the standard normal distribution table.
First, calculate the z-score:
z = (x - mean) / standard deviation
z = (2.40 - 2.04) / 0.25
z = 1.44
Using the z-score, we can find the probability from the standard normal distribution table. The probability of a sack weighing more than 2.40 pounds is 0.9251.
Comparing this probability with the given probability of 0.4251, we can see that the statement is FALSE.
13.) FALSE
To find the required sample size, we can use the margin of error formula:
n = (z * s / E)^2
Here, z represents the z-score for the desired confidence level, s represents the standard deviation, and E represents the margin of error.
Given that the standard deviation is assumed to be no larger than 10 minutes and the margin of error is ±2.00 minutes, we can use these values in the formula.
For a 90% confidence level, the z-score is approximately 1.645 (obtained from the standard normal distribution table).
Plugging in the values:
n = (1.645 * 10 / 2)^2
n = 41.125^2
n ≈ 1699.26
Therefore, the required sample size should be approximately 1699 students, not 60. Hence, the statement is FALSE.
9.) To determine if the statement is true or false, we need to calculate the z-score and find the corresponding probability.
The z-score formula is given by z = (x - μ) / σ, where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
In this case, we want to find the probability of a sack weighing more than 2.40 pounds, which means x = 2.40 pounds, μ = 2.04 pounds, and σ = 0.25 pounds.
The z-score can be calculated as follows:
z = (2.40 - 2.04) / 0.25 = 1.44
Now, we need to find the probability corresponding to this z-score using a standard normal distribution table or a z-table.
Looking up the value of z = 1.44 in the table, we find that the area to the left of this z-score is approximately 0.9251.
Since we want to find the probability of a sack weighing MORE than 2.40 pounds, we need to subtract this probability from 1 to get the area to the right of the z-score:
1 - 0.9251 = 0.0749
Therefore, the probability of a sack weighing more than 2.40 pounds is approximately 0.0749, not 0.4251.
Thus, the statement is FALSE.
13.) To determine if the statement is true or false, we need to calculate the sample size using the formula for estimating the sample size for a given confidence level and margin of error.
The formula to calculate the sample size is given by:
n = (z * σ / E) ^ 2
Where n is the sample size, z is the z-score corresponding to the desired confidence level (in this case, 90% confidence level), σ is the estimated standard deviation, and E is the margin of error.
In this case, the margin of error is ±2.00 minutes and the desired confidence level is 90%. However, we need to estimate the standard deviation, σ, to use in the formula.
Since the manager believes that the population standard deviation will not exceed 10 minutes, we can use this estimate.
Plugging in the values:
n = (z * σ / E) ^ 2 = (1.645 * 10 / 2) ^ 2
Calculating this:
n = (16.45/2) ^ 2 = (8.225) ^ 2 ≈ 67.75
Therefore, the sample size that should be used is approximately 67.75 students, not 60 students.
Thus, the statement is FALSE.