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1 year ago

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Last year Beth's annual salary was $38,350. This year she received a promotion and now earns $46,462 annually. She is paid biwee

kly.
a.) What was her biweekly salary last year?
b.) What is Beth’s biweekly salary this year?
c.) On a biweekly basis, how much more does Beth earn as a result of her promotion?

1 answer:

kenny6666 [7]1 year ago

8 0

Answer:

Quite simple. There are 52 weeks in a year. She gets paid 26 times, every other week.

38,350 / 26 = her salary from last year per check.

46,462 / 26 = her new salary per check

(46,462 - 38,350) / 26 = the difference per week

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A quality control manager at an auto plant measures the paint thickness on newly painted cars. A certain part that they paint ha

Scorpion4ik [409]

Answer:

78.88% probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 2, \sigma = 0.8, n = 100, s = \frac{0.8}{\sqrt{100}} = 0.08$

What is the probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value?

This is the pvalue of Z when X = 2 + 0.1 = 2.1 subtracted by the pvalue of Z when X = 2 - 0.1 = 1.9. So

X = 2.1

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

By the Central Limit Theorem

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

$Z = \frac{2.1 - 2}{0.08}$

$Z = 1.25$

$Z = 1.25$ has a pvalue of 0.8944

X = 1.9

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

$Z = \frac{1.9 - 2}{0.08}$

$Z = -1.25$

$Z = -1.25$ has a pvalue of 0.1056

0.8944 - 0.1056 = 0.7888

78.88% probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value

8 0

2 years ago

Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

Lina20 [59]

Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

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.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

a. at most 6 will come to a complete stop?

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

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

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133$

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$ . So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958$

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

$E(X) = np = 20*0.2 = 4$

4 of the next 20 drivers do you expect to come to a complete stop

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2 years ago

The average age of doctors in a certain hospital is 45.0 years old. Suppose the distribution of ages is normal and has a standar

Ad libitum [116K]

Answer: 0.7619

Step-by-step explanation:

Given : Mean : $\mu=45.0$

Standard deviation : $\sigma =8.0$

Sample size : $n=9$

We assume that the variable is normally distributed.

The value of z-score is given by :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

a) For x= 46.9 years

$z=\dfrac{46.9-45.0}{\dfrac{8}{\sqrt{9}}}=0.7125$

The p-value : $P(z$

Hence, the probability that the average age of those doctors is less than 46.9 years =0.7619

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2 years ago

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Triss [41]

Answer:

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Step-by-step explanation:

Both have y-intercepts at (0,1) but g(x) has a slope of 4 while f(x) has a slope of 5.

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