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marusya05 [52]
2 years ago
14

The scores on a statistics test had a mean of 81 and a standard deviation of 9. One student was absent on the test day and his s

core wasn't included in the calculation. If his score of 78 was added to the distribution of scores, what would happen to the mean and standard deviation?
A) Mean will increase, standard deviation will decrease.

B) Mean will increase, standard deviation will increase.

C) Mean will decrease, standard deviation will stay the same.

D) Mean will decrease, standard deviation will increase.

E) Mean will decrease, standard deviation will decrease.
Mathematics
1 answer:
Art [367]2 years ago
8 0

Answer: E) Mean will decrease, standard deviation will decrease

Step-by-step explanation:

Initial mean = 81

Initial standard deviation = 9

New score = 78

Taking a look at the initial mean(averahe) score, which is 81 and the new score to be added to the initial scores, the initial average is greater than the new score. Hence, this will result in a decrease in the new mean score after adding the new score of 78.

Also, taking a look at the standard deviation which is 9, we can conclude that the variability in initial scores from the mean scores is high. However, the new score of 78 and the initial mean are very close, and tend to low variation. Hence, adding the new score will lead to a decrease in variability and hence a decrease in the value of standard deviation.

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Let w represent the amount the wife receives.
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2 years ago
The amount of time a passenger waits at an airport check-in counter is random variable with mean 10 minutes and standard deviati
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Answer:

(a) less than 10 minutes

= 0.5

(b) between 5 and 10 minutes

= 0.5

Step-by-step explanation:

We solve the above question using z score formula. We given a random number of samples, z score formula :

z-score is z = (x-μ)/ Standard error where

x is the raw score

μ is the population mean

Standard error : σ/√n

σ is the population standard deviation

n = number of samples

(a) less than 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Therefore, the probability that the average waiting time waiting in line for this sample is less than 10 minutes = 0.5

(b) between 5 and 10 minutes

i) For 5 minutes

x = 5 μ = 10, σ = 2 n = 50

z = 5 - 10/2/√50

z = -5 / 0.2828427125

= -17.67767

P-value from Z-Table:

P(x<5) = 0

Using the z table to find the probability

P(z ≤ 0) = P(z = -17.67767) = P(x = 5)

= 0

ii) For 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Hence, the probability that the average waiting time waiting in line for this sample is between 5 and 10 minutes is

P(x = 10) - P(x = 5)

= 0.5 - 0

= 0.5

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Answer:

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Step-by-step explanation:

There is no attached diagram, but despite this, the tip amount can be calculated, since we know the total value of the expense in the restaurant and the percentage of tips they want to give, therefore it would be the multiplication of the total by the percentage like this:

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Answer:

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Answer:

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