Mathematics, 18.07.2019 17:00 rntaran2002

The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 8.3 cm. a. find the probability that an individual distance is greater than 214.30 cm. b. find the probability that the mean for 20 randomly selected distances is greater than 202.80 cm. c. why can the normal distribution be used in part (b), even though the sample size does not exceed 30? a. the probability is.

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Answer from: AkramMasoud

a. Find the probability that an individual distance is greater than 214.30 cm

We find for the value of z score using the formula:

z = (x – u) / s

z = (214.30 – 205) / 8.3

z = 1.12

Since we are looking for x > 214.30 cm, we use the right tailed test to find for P at z = 1.12 from the tables:

P = 0.1314

b. Find the probability that the mean for 20 randomly selected distances is greater than 202.80 cm

We find for the value of z score using the formula:

z = (x – u) / s

z = (202.80 – 205) / 8.3

z = -0.265

Since we are looking for x > 202.80 cm, we use the right tailed test to find for P at z = -0.265 from the tables:

P = 0.6045

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

I believe this is because we are given the population standard deviation sigma rather than the sample standard deviation. So we can use the z test.

Answer from: calebmoore925

Probability that an individual distance is greater than 214.30 cm is 0.1314 .

A)A -score is a numerical measurement that describes a value's relationship to the mean of a group of values.- -score is measured in terms of standard deviations from the mean

$z=\dfrac{\bar x-\mu}{\sigma }$

Where,

$\bar x=214.30 \\\mu=205\\\sigma=8.3$

Now put value in formula , we get

$z=\dfrac{214.30-205}{8.3}\\z=1.12$

So use the right tailed test to find Probability at z=1.12 from the tables,

We get Probability=0.1314 .

B) The probability that the mean for 20 randomly selected distances is greater than 202.80 cm is again find by using the forumla of - -score

$z=\dfrac{\bar x-\mu}{\sigma }$

where ,

$\bar x=202.80 \\\mu=205\\\sigma=8.3$

Now put value in formula , we get

$z=\dfrac{202.80-205}{8.3}\\\\z=-0.265$

So use the right tailed test to find Probability at z=-0.265 from the tables,

We get Probability=0..6045

C) We use Normal distribution in part b because we are given standard deviation rather instead of sample standard deviation. So the test can be used.

Answer from: Quest

well my understanding of this question is they want you to use the y=14.1x -2.3 and put the coordinates they gave you onto the line segment of y=14.1x -2.3

step-by-step explanation:

Answer from: Quest

d i think

step-by-step explanation:

Me plz i just need this one last test to pass

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