# Statistics #1&2 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...

Statistics #1&2

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?

### 3 Answers | Add Yours

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.

**Sources:**

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.

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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

`14*(2/3)~~ 9.33cm`

`9*(2/3) =6cm`

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`

**Sources:**