A car manufacturer produced 5,000 cars for a limited edition model. Dealers sold all of these cars at mean price

A large disaster cleaning company estimates that 30 percent of the jobs it bids on are finished within the bid time. Looking at

d1i1m1o1n [39]

Answer:

68.26% probability that the number of jobs finished on time is within 1 standard deviation of the mean.

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.

Looking at a random sample of 8 jobs that it has contracted, find the probability that x (number of jobs finished on time) is within one standard deviation of the mean.

Within 1 standard deviation of the mean is from Z = -1 to Z = 1. So this probability is the pvalue of Z = 1 subtracted by the pvalue of Z = -1.

Z = 1 has a pvalue of 0.8413

Z = -1 has a pvalue of 0.1587

So there is a 0.8413 - 0.1587 = 0.6826 = 68.26% probability that the number of jobs finished on time is within 1 standard deviation of the mean.

5 0

4 years ago

The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approxi

VikaD [51]

Answer:

$512.90 should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal 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.

Normally distributed with mean $480 and standard deviation $20.

This means that $\mu = 480, \sigma = 20$

How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05?

This is the 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95, so X when Z = 1.645.

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

$1.645 = \frac{X - 480}{20}$

$X - 480 = 1.645*20$

$X = 512.9$

$512.90 should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05

5 0

3 years ago

How about identify witch stamt is true A. -2.6>2.521 B.-0.4>-0.004 C.-8.0329>-8.045 D.-36.5>-36.07

astra-53 [7]

<span>A. -2.6 > 2.521 false
B.-0.4 > -0.004 false
C.-8.0329 > -8.045 true
D.-36.5 > -36.07 false

answer

</span>C.-8.0329 > -8.045

5 0

4 years ago

What is −45−−−−√ in simplest form?

emmasim [6.3K]

Answer: Option 2 or 3 times the square root of -5

7 0

3 years ago

How do you solve this equation ?

Lyrx [107]

Is z an angle? if so, i'll use the ration [email protected]=perpendicular/base
tanZ=15/20=3/4 answer

5 0

3 years ago