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baherus [9]
2 years ago
5

For a sample of 100 students, the median number of text messages sent per day is 90, the first quartile is 78, and the third qua

rtile is 102. Select all of the answers that create a true statement.
Approximately 25 students in the sample send

more than 90 text messages.
less than 90 text messages.
more than 78 text messages.
more than 102 text messages.
between 90 and 102 text messages.
less than 78 text messages.
between 78 and 102 text messages.
Mathematics
2 answers:
ohaa [14]2 years ago
5 0


Approximately 25 students send
Less than 78 text messages
More than 102 text messages
slavikrds [6]2 years ago
4 0

Answer with explanation:

Q={\text{Median}}=90\\\\Q_{1}=78\\\\Q_{3}=102

There are 100 students in the group.

If we divide 100 students into groups

→First quartile =25.5

means, first quartile of students has median value ≤ 78.

→Median =50.5

Most of the students sends ,90 messages a day.

→Third Quartile=75.5

Students lying above the third Quartile ,sends ≥102 messages per day.

First Quartile means, approximately 0- 25 students in the group , sends 78 messages.

Q_{1}\leq 78

Third Quartile means, 75-100 , students in the group, sends 102 messages .

Q_{3}\geq 102

Approximately 25 students in the sample send

1. more than 102 text messages.

2.less than 78 text messages.

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Step-by-step explanation:

You need to analize the information given in the exercise, You know that every time Jack fills the feeder, he put \frac{1}{3} pounds into it.

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3= \frac{x}{4}\\\\(3)(4)=x\\\\x=12

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Answer:

We conclude that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours.

Step-by-step explanation:

We are given that a sports statistician claim that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours.

The mean winning time of a sample of 30 randomly selected Boston marathon women's open division champions is 2.60 hours. assume the population standard deviation is 0.32 hours.

<em>Let </em>\mu<em> = </em><u><em>mean winning times for Boston marathon women's open division champions.</em></u>

So, Null Hypothesis, H_0 : \mu\geq 2.68 hours      {means that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours}

Alternate Hypothesis, H_A : p < 2.68 hours      {means that the mean winning times for Boston marathon women's open division champions is less than 2.68 hours}

The test statistics that would be used here <u>One-sample z test</u> <u>statistics</u> as we know about the population standard deviation;

                        T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean winning time = 2.60 hours

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Now, P-value of the test statistics is given by following formula;

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