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
iogann1982 [59]
3 years ago
8

Sick-leave time used by employees of a firm in a course of one month has approximately normal distribution, with a mean of 200 h

ours and a variance of 400 hours. a.Find the probability that total sick leave for next month will be less than 150 hours.b.In planning schedules for next month, how much time should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.
Mathematics
1 answer:
Usimov [2.4K]3 years ago
6 0

Answer:

a)0.62% probability that total sick leave for next month will be less than 150 hours.

b) 225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 200, \sigma = \sqrt{400} = 20

a.Find the probability that total sick leave for next month will be less than 150 hours.

This probability is the pvalue of Z when X = 150. So:

Z = \frac{X - \mu}{\sigma}

Z = \frac{150 - 200}{20}

Z = -2.5

Z = -2.5 has a pvalue of 0.0062.

So there is a 0.62% probability that total sick leave for next month will be less than 150 hours.

b.In planning schedules for next month, how much time should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

This is the value of X when Z has a pvalue of 0.90. So Z = 1.28

Z = \frac{X - \mu}{\sigma}

1.28 = \frac{X - 200}{20}

X - 200 = 20*1.28

X = 225.6

225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

You might be interested in
Triangle ABC is ransformed to obtain angle ABC
salantis [7]

Answer:How do you rotate 180 degrees about the origin?

Rotation by 180° about the origin:

The rule for a rotation by 180° about the origin is (x,y)→(−x,−y) .

Step-by-step explanation:

3 0
2 years ago
A mortgage is in the amount of $600,000. The origination fee is 0.5%, the intangible tax is 0.2%, and there are 1.5 discount poi
olchik [2.2K]
600,000×(0.005+0.002+0.015)
=13,200
4 0
3 years ago
What is the length of side x? *<br> 390<br> 18
Brrunno [24]
Uhh could you please include a image or something to help someone know it more Cleary with what you need help with
8 0
3 years ago
62% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are r
dedylja [7]

Answer:

a) 0.1180 = 11.80% probability that exactly 30 of them are spayed or neutered.

b) 0.8665 = 86.65% probability that at most 33 of them are spayed or neutered.

c) 0.4129 = 41.29% probability that at least 31 of them are spayed or neutered.

d) 0.5557 = 55.57% probability that between 24 and 30 of them are spayed or neutered.

Step-by-step explanation:

To solve this question, we use the binomial probability distribution, and also it's approximation to the normal distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

62% of owned dogs in the United States are spayed or neutered.

This means that p = 0.62

48 owned dogs are randomly selected

This means that n = 48

Mean and standard deviation, for the approximation:

\mu = E(x) = np = 48*0.62 = 29.76

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{48*0.62*0.38} = 3.36

a. Exactly 30 of them are spayed or neutered.

This is P(X = 30), which is not necessary the use of the approximation.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 30) = C_{48,30}.(0.62)^{30}.(0.38)^{18} = 0.1180

0.1180 = 11.80% probability that exactly 30 of them are spayed or neutered.

b. At most 33 of them are spayed or neutered.

Now we use the approximation. This is, using continuity correction, P(X \leq 33 + 0.5) = P(X \leq 33.5), which is the pvalue of Z when X = 33.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{33.5 - 29.76}{3.36}

Z = 1.11

Z = 1.11 has a pvalue of 0.8665

0.8665 = 86.65% probability that at most 33 of them are spayed or neutered.

c. At least 31 of them are spayed or neutered.

Using continuity correction, this is P(X \geq 31 - 0.5) = P(X \geq 30.5), which is 1 subtracted by the pvalue of Z when X = 30.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{30.5 - 29.76}{3.36}

Z = 0.22

Z = 0.22 has a pvalue of 0.5871

1 - 0.5871 = 0.4129

0.4129 = 41.29% probability that at least 31 of them are spayed or neutered.

d. Between 24 and 30 (including 24 and 30) of them are spayed or neutered.

This is, using continuity correction, P(24 - 0.5 \leq X \leq 30 + 0.5) = P(23.5 \leq X \leq 30.5), which is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 23.5.

X = 30.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{30.5 - 29.76}{3.36}

Z = 0.22

Z = 0.22 has a pvalue of 0.5871

X = 23.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{23.5 - 29.76}{3.36}

Z = -1.86

Z = -1.86 has a pvalue of 0.0314

0.5871 - 0.0314 = 0.5557

0.5557 = 55.57% probability that between 24 and 30 of them are spayed or neutered.

8 0
3 years ago
The density function for the number of times the riders scream on a roller coaster is given by....
larisa [96]
W. w w w q w. Q q q q. Q q
3 0
3 years ago
Other questions:
  • The ratio opposite​/adjacent describes which function?
    11·1 answer
  • Write in graphing form please and please show the work!! y= x^2 +2x-3
    6·1 answer
  • What is 4.76 hours converted to min and sec
    10·2 answers
  • After you measure the dimensions of a plot of land with a steel tape on a hot day, you come back and re-measure on a cold day. O
    11·1 answer
  • Find the slope of the line through the points (3,2) (8,6).
    12·1 answer
  • Explain why a positive discriminant results in two real solutions.
    11·1 answer
  • What is the volume of a cylinder that has the diameter of 8.5 inches and a height of 6.4 inches?
    15·1 answer
  • What is 3a + 3/6. if a = 1/4
    8·1 answer
  • Point B: (14,4)Point A: (5,4)What is the difference between the x-coordinate of point A and the x-coordinate of point B?
    15·1 answer
  • Zero times 1 million is zero
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!