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jonny [76]
3 years ago
7

The Community College Survey of Student Engagement reports that 46% of the students surveyed rarely or never use peer or other t

utoring resources. Suppose that in reality 40% of community college students never use tutoring services available at their college. In a simulation, we select random samples from a population in which 40% do not use tutoring. For each sample, we calculate the proportion who do not use tutoring. If we randomly sample 100 students from this population, the standard error is approximately 5%. Would it be unusual to see 46% who do not use tutoring in a random sample of 100 students?a) Yes, this would be unusual because 46% is more than one standard error from the mean. It is very rare for a sample to be more than one standard error from mean. b) Yes, this would be unusual because 46% is 6% higher than 40%. c) No, this would not be unusual because the error is only 6%. d) No, this would not be unusual because 46% is only 1.2 standard errors from 40%.
Mathematics
1 answer:
kap26 [50]3 years ago
3 0

Answer:

d) No, this would not be unusual because 46% is only 1.2 standard errors from 40%.

Step-by-step explanation:

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.

If the z-score is higher than 2 or lower than -2, X is unusual.

In this question:

Mean = 40%. So \mu = 0.4

Standard error = 5%. So \sigma = 0.05

Is 46% unusual?

We have to find Z when X = 0.46. So

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

Z = \frac{0.46 - 0.4}{0.05}

Z = 1.2

1.2 is lower than 2, that is, it is only 1.2 standard deviations from the mean.  So 46% is not unusual.

So the correct answer is:

d) No, this would not be unusual because 46% is only 1.2 standard errors from 40%.

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Un jardinier a récolté deux quintaux trois quarts de pommes . Il vend un quintal un quart à un voisin ; il vend aussi 7/10 de qu
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Answer:

Step-by-step explanation:

Notez que: un quintal =

une unité de poids égale à 100 kg

Un jardinier a récolté deux quintaux et trois quarts de pommes.

Le nombre total de pommes récoltées = 200 kg + 3/4 du quintal de pommes

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Il vend un quintal un quart à un voisin

Le montant qu'il a vendu à un voisin = 100 + 1/4 quintal de pommes

100 + 0,127

= 100,127 kg

Il vend également 7/10 d'un quintal sur le marché du village

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= 70 kg de pommes restantes

2/5 d'un quintal à un pâtissier.

2/5 × 100 kg

= 40 kg de pommes restantes

À l'aide d'une fraction, exprimez la quantité de pomme que le jardinier a gardée pour lui.

La quantité que le jardinier lui a réservée =

200 3/4 kg - (100 1/4 kg + 70 kg + 40 kg)

200,381 kg - (210,127 kg)

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The function y=5x+10 represents the total cost y of a group of x members to go paintballing.
never [62]

Answer:

The independent variable (x) represents the members to go paintballing.

The dependent variable (y) represents the total cost.

Step-by-step explanation:

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