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rusak2 [61]
3 years ago
10

A bus makes a stop at 2:30 it's letting off 15 people and letting on 9 the bus makes another stop 10 minutes later to let off 4

more people how does the number of people on the bus after the second stop compare the number of people on the bus before the 2:30 it stop?
Thank you for answering!
Mathematics
2 answers:
Luden [163]3 years ago
6 0
It changes by 10 people.
dsp733 years ago
5 0
X-15+9-4=x-10 there are ten fewer people after the second stop than the original amount
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The mean number of words per minute (WPM) read by sixth graders is 97 with a standard deviation of 19 WPM. If 75 sixth graders a
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Answer:

0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM

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 = 97, \sigma = 19, n = 75, s = \frac{19}{\sqrt{75}} = 2.1939

What is the probability that the sample mean would be greater than 101.63 WPM?

This is 1 subtracted by the pvalue of Z when X = 101.63. So

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

By the Central Limit Theorem

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

Z = \frac{101.63 - 97}{2.1939}

Z = 2.11

Z = 2.11 has a pvalue of 0.9826.

1 - 0.9826 = 0.0174

0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM

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