How do you put the answer in a excel format
These z values can be obtained using Excel’s NORM.S.INV function or by using the standard normal probability table in the text. a. The z value corresponding to a cumulative probability of .2119 is z = -.80. NORM.S.INV(.2119) b. Compute .9030/2 = .4515; z corresponds to a cumulative probability of .5000 + .4515 = .9515. So z = 1.66. NORM.S.INV(.9515) c. Compute .2052/2 = .1026; z corresponds to a cumulative probability of .5000 + .1026 = .6026. So z = .26. NORM.S.INV(.6026) d. The z value corresponding to a cumulative probability of .9948 is z = 2.56. NORM.S.INV(.9948) e. The area to the left of z is 1 - .6915 = .3085. So z = -.50. NORM.S.INV(.3085)
| Probability | z-value |
| 0.2119 | -0.80 |
| 0.9515 | 1.66 |
| 0.6026 | 0.26 |
| 0.9948 | 2.56 |
| 0.3085 | -0.50 |
Go to excel > Home> format as table >select the table(and Range).
now Rename column as Probability and z value respectively
enter probability values in Probability column, in corresponding z-value column use formula =NORMSINV(probability) it will automatically generate all the probabilities.
How do you put the answer in a excel format These z values can be obtained...
also if you know how to do it through excel, can you please include
steps? thank you !
Suppose a sample of O-rings was obtained and the wall thickness (in inches) of each was recorded Use a normal probability plot to 0.165 0.191 0.194 0 2195 assess whether the sample data could have come from a population that is normally distributed 0.216 0236 0.233 0.240 0.248 0 261 0 272 0.283 0.282 0.308 0.310 0334 Click here to view the...
Use Excel Functions and show which functions you used. You may
take screenshots.
For Problems #1 through #9, consider the following information: According to product packaging, the calorie count for a single house brand premium milk chocolate chip pecan cookie baked exclusively for Sale Mart is 200 calories. A consumer health testing lab tech is double checking this claim, and has collected a random sample of 16 individual milk chocolate chip pecan cookies. The corresponding calories were as follows 193...
Let X be a normal random variable with mean 4 and variance 9. Use the normal table to find the following probabilities, to an accuracy of 4 decimal places. Normal Table The entries in this table provide the numerical values of Φ(z)=P(Z≤z), where Z is a standard normal random variable, for z between 0 and 3.49.For example, to find Φ(1.71), we look at the row corresponding to 1.7 and the column corresponding to 0.01, so that Φ(1.71)=.9564. When z is...
help please!
ignore my work as i tried to get the answers!
Find the z score if the NON-shaded region indicated by the arrow is .0991 Round to 4 decimal places -.5.0991 2.4009 25 0991 Applicable z score(s) Final probability as a 4 place decimal IF APPLICABLE Movie theater managers measured the number of scenes in Scary Movie that produced laughter and found the average to be 82.1 separate scenes with stdev =13.06 scenes. Statisticians tested adqiç audiences and found...
The average annual cost (incduding tuition, room, board, boeks and fees) to attend a public college takes nearly a third of the annual income of a typical family with college-age children of attending private and public colleges. Data are in thousands of dollars. Click on the datafile logo to reference the data s the ana al cost O Money, April 2012 At private colleges, the average annual cost is equal to about 60% of the typical famly's income. The tolle...
A survey found that women's heights are normally distributed with mean 63.7 in and standard deviation 2.3 in. A branch of the military requires women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all...
Name: Week 7 HSCI 390L: Hypothesis Testing Worksheet 1. It has been reported that the average credit card debt for college seniors is $3262. The student senate at a large university feels that their seniors have a debt much less than this. so they conduct a study of 50 randomly selected seniors and finds that the average debt is $2995, and the population standard deviation is $1100. Let's conduct the test based on an alpha level of 0.05. Complete the...
t-Distribution Table Find the critical value t, for the confidence level c0 80 and sample size n 9 Click the icon to view the t-distribution table. (Round to the nearest thousandth as needed ) Level of confidence, c 0.80 0.90 0.95 0.98 0.9 One tall a 0.10 0.05 0.025 0.01 0.05 d. Two talls, α 0.20 0.10 005 002 001 du 3.078 6 314 12.706 31.821 63.657 1886 2920 4303 6965 9925 2 638 2.353 3.182 4541 5.841 1.533 2.132...
I ONLY NEED HELP WITH PART OF PART "B"
I've figured out the test statistic is -1.73 and the degrees of
freedom are 5. However, I'm having a hard time finding the P value
via the chart (which I'm required to learn how to do).I think the
chart immediately bellow this is the one used to find the p-value.
However, I know at least one (or more) of the charts bellow is
what's used. Please let me know which chart...
I ONLY NEED HELP WITH PART OF PART "B"
I've figured out the test statistic is -1.73 and the degrees of
freedom are 5. However, I'm having a hard time finding the P value
via the chart (which I'm required to learn how to do).I think the
chart immediately bellow this is the one used to find the p-value.
However, I know at least one (or more) of the charts bellow is
what's used. Please let me know which chart...