The average playing time of albums in a certain collection is 27 minutes, with a standard deviation of 4 minutes. Calculate the z-score, to the nearest thousandth, for an album that has a playing time of 29.5 minutes.(1 point)
z=
0.625
Estimate the area percentage under a normal curve to the left of a z-score of −0.4
.(1 point)
percent
The area percentage under a normal curve to the left of a z-score of -0.4 can be found using a standard normal distribution table, which gives the percentage of values to the left of a given z-score.
Looking up the z-score of -0.4 in the table, we find that the area to the left of -0.4 is approximately 0.3446 or 34.46%.
Therefore, the estimated area percentage under the normal curve to the left of a z-score of -0.4 is 34.46%.
A factory makes aluminum cans that will be filled with soda. The walls of the cans have a mean thickness of 0.1 mm and a standard deviation of 0.02 mm. What is the probability that a randomly selected can will have walls that are less than 0.118 mm in thickness?
(1 point)
The probability that the thickness of the walls of a randomly selected can will be less than 0.118 mm is
percent.
To solve this problem, we need to standardize the value of 0.118 mm using the given mean and standard deviation:
Z = (X - μ) / σ
Z = (0.118 - 0.1) / 0.02
Z = 0.018 / 0.02
Z = 0.9
Now, we need to find the probability that a randomly selected can will have walls that are less than 0.118 mm, which is equivalent to finding the probability of Z < 0.9 from the standard normal distribution table.
Using the standard normal distribution table or calculator, the probability of Z < 0.9 is approximately 0.8159 or 81.59%.
Therefore, the probability that the thickness of the walls of a randomly selected can will be less than 0.118 mm is 81.59%.
The mean height of adult males in the United States is 69 inches with a standard deviation of 2.5 inches. What is the probability that a randomly-selected adult male will have a height between 69.3 inches and 71.2 inches?
(1 point)
Responses
26.28%
26.28%
84.13%
84.13%
73.72%
73.72%
77.64%
To find the probability that a randomly-selected adult male will have a height between 69.3 inches and 71.2 inches, we need to first standardize these values using the mean and standard deviation given:
Convert 69.3 inches and 71.2 inches to Z-scores:
Z1 = (69.3 - 69) / 2.5 = 0.12
Z2 = (71.2 - 69) / 2.5 = 0.88
Next, we look up the probabilities corresponding to these two Z-scores in the standard normal distribution table:
P(Z < 0.12) = 0.4515
P(Z < 0.88) = 0.8106
To find the probability of the range between 69.3 inches and 71.2 inches, we subtract these probabilities:
P(69.3 < X < 71.2) = P(69 < X < 71.2) = P(Z < 0.88) - P(Z < 0.12) = 0.8106 - 0.4515 = 0.3591.
Then we convert this back to percentage:
0.3591 * 100% = 35.91%.
Therefore, the probability that a randomly-selected adult male will have a height between 69.3 inches and 71.2 inches is approximately 35.91%.
Use the table to answer the question.
The top-left cell of a z-score table is labeled z. The remaining cells in the header row have values ranging from 0.00 to 0.09 in increments of 0.01. Below z, the first column has values ranging from 0.3 to 2.1 in increments of 0.1.
The mean height of adult males in the United States is 69 inches with a standard deviation of 2.5 inches. What is the probability that a randomly-selected adult male will have a height between 69.3 inches and 71.2 inches?
(1 point)
Responses
26.28%
26.28%
84.13%
84.13%
73.72%
73.72%
77.64%
To find the probability that a randomly-selected adult male will have a height between 69.3 inches and 71.2 inches, we first need to calculate the Z-scores for these heights using the given mean and standard deviation.
Z1 = (69.3 - 69) / 2.5 = 0.12
Z2 = (71.2 - 69) / 2.5 = 0.88
Next, we need to find the corresponding probabilities for these Z-scores from the standard normal distribution table. Look for the value closest to 0.12 in the z-column and 0.88 in the 0.08 row.
From the table, the probabilities for Z = 0.12 and Z = 0.88 are approximately 0.5478 and 0.8119, respectively.
To find the probability of the range between 69.3 inches and 71.2 inches, we subtract these probabilities:
P(69.3 < X < 71.2) = P(69 < X < 71.2) = P(Z < 0.88) - P(Z < 0.12) = 0.8119 - 0.5478 = 0.2641.
Converting this probability to a percentage:
0.2641 * 100% = 26.41%.
Therefore, the probability that a randomly-selected adult male will have a height between 69.3 inches and 71.2 inches is approximately 26.41%.
The closest response option is 26.28%.