1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
grin007 [14]
3 years ago
13

The overhead reach distances of adult females are normally distributed with a mean of 205.5 and a standard deviation of 8.6 . a.

Find the probability that an individual distance is greater than 218.00 cm. b. Find the probability that the mean for 15 randomly selected distances is greater than 204.00 c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?
Mathematics
1 answer:
Anastasy [175]3 years ago
3 0

Answer:

a) 0.073044

b) 0.75033

c) The normal distribution can be used in part (b), even though the sample size does not exceed 30 because initial population size is been distributed normally , therefore, the mean of the samples will be normally distributed regardless of their size(meaning whether the sample size is less than or equal to or exceeds 30, the sample means the mean of the samples will be normally distributed regardless of their size).

Step-by-step explanation:

The overhead reach distances of adult females are normally distributed with a mean of 205.5 and a standard deviation of 8.6 .

a. Find the probability that an individual distance is greater than 218.00 cm.

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = 218

μ is the population mean = 205.5

σ is the population standard deviation = 8.6

For x > 218

z = 218 - 205.5/8.6

z = 1.45349

Probability value from Z-Table:

P(x<218) = 0.92696

P(x>218) = 1 - P(x<218) = 0.073044

b. Find the probability that the mean for 15 randomly selected distances is greater than 204.00

When = random number of samples is given, we solve using this z score formula

z = (x-μ)/σ/√n

where

x is the raw score = 204

μ is the population mean = 205.5

σ is the population standard deviation = 8.6

n = 15

For x > 204

Hence

z = 204 - 205.5/8.6/√15

z = -0.67552

Probability value from Z-Table:

P(x<204) = 0.24967

P(x>204) = 1 - P(x<204) = 0.75033

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The normal distribution can be used in part (b), even though the sample size does not exceed 30 because initial population size is been distributed normally , therefore, the mean of the samples will be normally distributed regardless of their size(meaning whether the sample size is less than or equal to or exceeds 30, the sample means the mean of the samples will be normally distributed regardless of their size).

You might be interested in
The total bill for the drinks and pizza comes out to be $17.28. How much should each person pay if they are to share the total b
sleet_krkn [62]

Answer:

5.76

Step-by-step explanation:

17.28/3

5 0
2 years ago
Read 2 more answers
-5(m+4)-10(8-5m) which’s is this
Darina [25.2K]
-5m-20-80+50m=0
45m=100
m= 20/9
3 0
3 years ago
Read 2 more answers
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
Oksana_A [137]

Answer:

n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

 In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

{n \choose 5 } = \frac{n!}{5!(n-5)!}

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 {n \choose 4} .

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is 4 * {n \choose 4} .

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is {n \choose 3} .

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have {3 \choose 2 } = 3  possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of 6 * {n \choose 3}

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of {n \choose 2}  possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.

The total amount of possibilities for this case is

4 * {n \choose 2}

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}

I hope that works for you!

4 0
3 years ago
Are the expressions x + x + 1 + x + 2 + x + 1 + x and 5x + 4 equivalent? Why or why not?
egoroff_w [7]

Answer:

yes

they are equivalent because lets say the X = 3 then you would have to answer

3 + 3 + 1 + 3 + 2 + 3 + 1 + 3

and that would = 19

and then on the other equation

5 x 3 + 4 = 19

So the both equal the same.

PLZ MARK ME BRAINLIEST

3 0
3 years ago
Kaley cuts half of a loaf of bread into 4 equal parts what fractons of the whole loaf dose each of the 4 parts reperzent
saveliy_v [14]
1/4 years s the answer because if you divide a whole in four you get 1/4
5 0
3 years ago
Read 2 more answers
Other questions:
  • Mu is walking laps to raise money for charity. For each lap she walks, her sponsors will donate $7. Mu has walked l laps and rai
    15·2 answers
  • Figure ABCD ≅ Figure MNOP.
    5·1 answer
  • Evaluate the expression: –(31 + 2) + 72 – (–5)2.
    10·1 answer
  • A. 12<br> B. 10<br> C. 6<br> D. 5
    8·1 answer
  • What is the range of y=-3sine(x)-4​
    6·1 answer
  • An open box is to be made from a flat square piece of material 19 inches in length and width by cutting equal squares of length
    6·1 answer
  • Help please and thank you: The graph of f x = x2 − 2x − 3 is shown. Which of these describes the effect changing f x to f −2x wi
    7·1 answer
  • The altitude to the hypotenuse of a right triangle divides the hypotenuse into segments 2 cm and 8 cm long. ​Find the length of
    5·1 answer
  • On a number line what nomber is 3 units to the left of 3​
    10·1 answer
  • Which of the following is a linear function?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!