Answer:
they would be $42.00
Step-by-step explanation:
50 + 2.50= 52.50
52.50-10.50=42.00
Answer: Choice B (If you need to do questions like this in the future i recommend using the desmos graphing calculator online, inputting the equations and comparing them)
Answer:
0.2857
Step-by-step explanation:
The probability of interest is the area under the probability density curve between the z-values associated with the temperature limits of interest.
__
The first attachment shows a "sketch" of the distribution and the area of the portion of interest. (It also shows the probability as 0.2858.)
__
The second attachment shows the table values of interest for this problem. The z-values that we want to look for in the table are ...
z1 = (0.50° -0°)/1.00° = 0.50
and
z2 = (2.00° -0°)/1.00° = 2.00
The area of the probability density function to the left of each of these z-values is given in the table, so the area between them is the difference of table values:
0.9772 -0.6915 = 0.2857
The probability of a reading between 0.50 and 2.00 is about 0.2857.
Answer:
- Mean will Increase .
- Median remains unchanged.
- Standard deviation will increase.
Step-by-step explanation:
We are given that there are 14 employees in a particular division of a company and their salaries have a mean of $70,000, a median of $55,000, and a standard deviation of $20,000.
And also the largest number on the list is $100,000 but By accident, this number is changed to $1,000,000.
Now we have to analyse the Effect of this change in data values on mean, median, and standard deviation.
- Mean will get affected because $1,000,000 is a very huge value as compared to $100,000 and is considered to be an outlier and we know that mean is affected by outliers as mean will change to $134285.7143 after replacing $100,000 with $1,000,000 .
- Median will not get affected as median the middle most value in the data set and since $1,000,000 is considered to be an outlier so median remain unchanged at $55,000 .
- Standard Deviation will also get affected as due to outlier value in the data set the numerator value will increase very much and due to which standard deviation will also increase.
Answer:
a) 229 and 305 days
b) 229 days or less
c) 305 days or more
Step-by-step explanation:
The Empirical Rule(68-95-99.7 rule) states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 267
Standard deviation = 19
(a) Between what values do the lengths of the middle 95% of all pregnancies fall?_____________and___________days
By the Empirical rule, 95% of all pregnancies fall within 2 standard deviations of the mean.
So
267 - 2*19 = 229 days
to
267 + 2*19 = 305 days
(b) How short are the shortest 2.5% of all pregnancies?______days or less
95% of all pregnancies fall within 2 standard deviations of the mean. The other 5% are more than 2 standard deviations from the mean. Since the distribution is symmetric, 2.5% is more than 2 standard deviations below the mean(shortest 2.5%) and 2.5% is more than 2 standard deviations above the mean(longest 2.5%). So
267 - 2*19 = 229 days
c) How long do the longest 2.5% of pregnancies last?________days or more
Explanation in b)
267 + 2*19 = 305 days
