can you show me the calculation ;)
The life expectancy of computer terminals is normally distributed with a mean of 4 years and standard deviation for 10 months.
a. What is the probability that a randomly selected terminal will last more than 5 years?
b. What percentage of terminals will last between 2.5 years and 4.5 years?
c. 80 percent of terminals will last at least how many years?
Convert all values either into months or years.
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. To get percentages, multiply by 100.
c. Reverse process to find Z score and calculate.
0.102
Sure! Let's solve each part step-by-step:
a. To find the probability that a randomly selected terminal will last more than 5 years, we need to convert the given values into standard units and then use the standard normal distribution table.
First, we need to convert 5 years into months. Since there are 12 months in a year, 5 years is equal to 60 months.
Next, we calculate the z-score using the formula:
z = (x - mean) / standard deviation
where x is the value we want to find the probability for, mean is the mean of the distribution, and standard deviation is the standard deviation of the distribution.
In this case, the mean is 4 years, which is equal to 48 months, and the standard deviation is 10 months.
z = (60 - 48) / 10
z = 12 / 10
z = 1.2
Using the standard normal distribution table, we can find the probability associated with a z-score of 1.2. Looking up this value in the table, we find that the area to the left of the z-score is approximately 0.8849.
Since we want to find the probability that a terminal will last MORE than 5 years, we subtract this value from 1:
P(x > 5 years) = 1 - 0.8849
P(x > 5 years) = 0.1151
Therefore, the probability that a randomly selected terminal will last more than 5 years is approximately 0.1151 or 11.51%.
b. To find the percentage of terminals that will last between 2.5 years and 4.5 years, we need to calculate the z-scores for those two values and again use the standard normal distribution table.
For 2.5 years, which is equal to 30 months:
z1 = (30 - 48) / 10
z1 = -18 / 10
z1 = -1.8
For 4.5 years, which is equal to 54 months:
z2 = (54 - 48) / 10
z2 = 6 / 10
z2 = 0.6
Using the standard normal distribution table, we can find the areas associated with z-scores of -1.8 and 0.6.
Looking up the value associated with z1 = -1.8, we find that the area to the left of this z-score is approximately 0.0359.
Looking up the value associated with z2 = 0.6, we find that the area to the left of this z-score is approximately 0.7257.
To find the percentage between these two values, we subtract the smaller area from the larger area:
P(2.5 years < x < 4.5 years) = 0.7257 - 0.0359
P(2.5 years < x < 4.5 years) = 0.6898
Therefore, approximately 68.98% of terminals will last between 2.5 years and 4.5 years.
c. To find the number of years that 80% of terminals will last at least, we need to find the corresponding z-score for the cumulative probability of 0.80 using the standard normal distribution table.
Looking up the value associated with a cumulative probability of 0.80, we find that the z-score is approximately 0.8416.
Now, we can use the z-score formula to find the corresponding x-value (number of months):
z = (x - mean) / standard deviation
0.8416 = (x - 48) / 10
Solving for x, we get:
0.8416 * 10 = x - 48
8.416 = x - 48
x = 48 + 8.416
x = 56.416
Converting this value back to years, we get:
56.416 months = 4.7013 years
Therefore, 80% of terminals will last at least approximately 4.7013 years.
Sure! To calculate the probabilities for these questions, we will use the standard normal distribution table (also known as the Z-table), which provides the probability values for the standard normal distribution.
First, let's convert the given values into the standardized form using the formula:
Z = (X - μ) / σ
Where:
Z is the standardized value,
X is the original value,
μ is the mean, and
σ is the standard deviation.
a. To find the probability that a randomly selected terminal will last more than 5 years, we need to calculate the Z-score for the value of 5 years.
Z = (5 - 48) / 10 = -0.4
Using the Z-table, we can find the probability corresponding to the Z-score of -0.4. The table provides the probability to the left of the Z-score. Since we want the probability of the terminal lasting more than 5 years, we subtract this value from 1.
The probability of a randomly selected terminal lasting more than 5 years is 1 - P(Z < -0.4).
b. To find the percentage of terminals that will last between 2.5 years and 4.5 years, we need to calculate the Z-scores for both values.
For 2.5 years:
Z1 = (2.5 - 48) / 10
For 4.5 years:
Z2 = (4.5 - 48) / 10
Using the Z-table, we can find the probabilities corresponding to both Z1 and Z2. The probability of terminals lasting between 2.5 years and 4.5 years is given by P(Z1 < Z < Z2).
c. To find the number of years that corresponds to 80 percent of terminals lasting at least that long, we need to find the Z-score that corresponds to the 80th percentile. The Z-score can be found by looking up the cumulative probability of 0.80 in the Z-table.
Once we have the Z-score, we can use the formula to solve for X:
X = Z * σ + μ,
Where X is the number of years, Z is the Z-score, σ is the standard deviation, and μ is the mean.
Using these calculations, you can find the answers to the three questions.