1: In population, 50% of the adults are taller than 172cm and 10% are taller than 190cm. determine the mean height and standard deviation for this population.
2: use confidence intervals to interpret the following statement. A recent study by a fitness centre indicated that members spend an averge of 216 min per week at the gym the results are considered accurate with a margin of error of -+2.8%, 9 times out 10.
3- A photograph from the internet that measures 14cm by 9 cm is reduced by scale factor of 2/3 so it can fit on a brochure.
a. what are dimensions of brochure image?
b. by what scale factor, to the nearest hundredth, was the area of the photograph decreased in the reduction process?
For question 1, you can safely assume that height in a population is normally distributed. In a normal distribution, the mean, median, and mode are the same.
So ` mu=172 ` .
If we then standardize the scores, a score with 10% of the population above it (equivalent to 90% below it) relates to a `z-"score" ` of 1.28. Since `z=(x-mu)/sigma ` we can compute the standard deviation `sigma ` by:
`1.28=(190-172)/sigma ==> sigma~~14.06"cm" ` .
For question 2, we can say with 90% confidence that the mean number of minutes spent at the fitness center is in the interval `209.95<=mu<=222.05 `
-- 90% confidence level comes from the statement 9 out of 10 times
-- We find the interval by adding and subtracting 2.8% of 216 minutes.
For question 3(b) note that if the scale factor between two similar objects is a:b, then corresponding areas of the two objects have a ratio of `a^2:b^2 ` . Since the scale factor is 2:3, the ratio of areas is 4:9.
Note that this generalizes to volumes as well. The ratio of corresponding volumes would be `a^3:b^3 `
Question 3(a) is simply a matter of computing the new sizes:
`14*2/3=9.bar(3),9*2/3=6 ` so the dimensions of the image are ` ``(28/3"cm")"x"(6"cm") `
See that `(28/3*6)/(14*9)=56/126=4/9 ` as computed above.
50% of adults are taller than 172 cm and 10% are taller than 190 cm. Nothing is known about the height of the other 50%. The information provided is insufficient to determine the mean height or the standard deviation for the population.
The photograph from the internet measures 14 cm by 9 cm. It is reduced by a scale factor of 2/3. After the image size has been reduced its length is 14*(2/3) ~~ 9.33 cm and the width is 9*(2/3) = 6 cm.
The dimension of the photograph after reduction is 9.33 cm by 6 cm.
The area of the original photograph was 14*9 = 126 cm^2. The area of the reduced image is (28/3)*6 = 56 cm^2. The scale factor is 4/9.
whaty about the question number #2 ?
1. Not enough information since you're missing data on the other 40% of the population.
2. A few things are needed to calculate the confidence interval, but all the values needed are already in the question, but I'll go through it anyways.
Basically, confidence interval = mean `+-` margin of error
The average (sometimes called the mean) is 216min.
The margin of error is `+-` 2.8.
The phrase "9 times out of 10" is the desired confidence level, which is used to calculate the margin of error (which is given in the question so we don't need it)
So putting everything together you'd have 216 `+-` 2.8
3a. Multiply the dimensions by 2/3 to get the new dimensions
Brochure image is 9.33cm x 6cm
3b. To find the scale factor, find the original area and then the reduced area. Then divide the two to get the scale.
Original Area: `14*9 = 126cm^2`
Reduced Area: `(14*(2/3)) * 6 = 56cm^2` (Note: Use the exact value for an accurate answer)
So the scale factor is `56/126 = 4/9`