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MissTica
4 years ago
5

1) Psychology students at Wittenberg University completed the Dental Anxiety Scale questionnaire. Scores on the scale range from

0 (no anxiety) to 20 (extreme anxiety). The mean score was 11 and the standard deviation was 3.5. Assume that the distribution of all scores on the dental anxiety scale is normal with with \mu = 11 and \sigma = 3.5.
a) Find the probability that someone scores between a 10 and a 15 on the Dental Anxiety Scale.
b) Find the probability that someone scores above a 17 on the Dental Anxiety Scale.
2) Ecological Applications published a study on the development of forests following wildfires in the Pacific Northwest. One variable of interest to the researcher was tree diameter at breast height 110 years after the fire. The population of Douglas fir trees was shown to have an approximately normal diameter distribution, with \mu = 50 centimeters and \sigma = 12 cm. Find the diameter, d, such that 30% of the Douglas fir trees in the population have diameters that exceed d.
Mathematics
1 answer:
Zarrin [17]4 years ago
8 0

Answer:

1) 0.487,0.0436

2) 56.3 cm

Step-by-step explanation:

Let X be the scores ranging from 0 to 20 of the  Psychology students at Wittenberg University completed the Dental Anxiety Scale questionnaire.

X is N(11,3.5)

a) the probability that someone scores between a 10 and a 15 on the Dental Anxiety Scale.

=P(10

b) P(X>17) = P(Z>1.71)\\= 0.5-0.4564\\=0.0436

c) Y is N(50, 20) where Y is tree diameter at breast height 110 years after the fire.

P(X>d) = 0.30

Corresponding Z value = 0.525

X= 50+12(0.525)\\=56.3

d = 56.3 cm

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Help plz!!!!!!!!!
SIZIF [17.4K]

Answer:

The first quartile is 61 and the third quartile is 63.

Step-by-step explanation:

From the given dot plot is noticed that

Temperature                  frequency            

   59                                   0

   60                                   2

   61                                    3

   62                                   3

   63                                   4

   64                                   2

   65                                   1

 Total                                 15

It means the data is in the form of

60,60,61,61,61,62,62,62,63,63,63,63,64,64,65

The median of data is middle term. The total number terms is 15. Therefore 7th terms is the median

(60,60,61,61,61,62,62),62,(63,63,63,63,64,64,65)

Now, we have 7 ters before median and 7 terms after median.

First quartile is the middle term of minimum value and median and third quartile is the middle term of maximum value and median.

So, 4th term is first quartile and 12th term is third quartile.

(60,60,61),61,(61,62,62),62,(63,63,63),63,(64,64,65)

Therefore the first quartile is 61 and the third quartile is 63.

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