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STAT 1 Assignment ,

7 Questions….

Needs to be finished correctly and ASAP please

(1) For the data below:

76, 48, 93, 60, 93, 65, 45, 80, 77, 80, 93, 88, 43, 44, 68, 89, 52, 68, 84, 78

(a) (5 pts) Construct a relative-frequency distribution by using limit grouping with a first class of 40-49 and a class width of 10.

(b) (5 pts) Construct a relative-frequency histogram for the frequency distribution in part (a).

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(2) (5 pts) Without using the built-in function for mean and standard deviation on your calculator, find the mean and standard deviation for the following sample data:

{17, 9, 11, 15, 3}. (Round the results to the nearest hundredth if applicable.)

(3) For the date set below:

{ 4, 10, 12, 12, 14, 14, 15, 16, 18, 20, 20, 21, 23, 25, 27, 36 } (a) (3 pts) Find the first, second and third quartiles of these data.

(b) (2 pts) Find the upper and lower limits of the data set.

(c) (2 pts) Construct a boxplot for the data.

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(4) (2 pts each) From the set S = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19}, a single number is to be selected randomly.

Let A be the event that the selected number is a multiple of 3.

Let B be the event that the selected number is a multiple of 2

(a) Find P(A).

(b) Find P(B).

(c) Find P(A & B).

(d) Find P(A or B).

(e) Find P(B|A). (f) Find P(A|B).

(5) (5 pts) According to the General Social Survey conducted at the University of Chicago, 59% of employed adults believe that if they lost their job, it would be easy to find another one with a similar salary. Suppose that 5 employed adults are randomly selected. Find the probability that more than 3 them believe it would be easy to find another job. (Round the result to the nearest thousandth if applicable.)

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(6) (5 pts) A bottler of drinking water fills plastic bottles with a mean volume of 990 milliliters (mL) and standard deviation 6 mL. The fill volumes are normally distributed. Determine the percentage of all bottles that have volumes greater than 987 mL.

(7) (5 pts) Eight measurements were made of the mineral content (in percent) of spinach, with the following results.

19.1, 20.8, 20.8, 21.4, 20.5, 19.7, 21.0, 19.2

(Sample mean: 20.3; sample standard deviation: 0.9)

It is reasonable to assume that the population is approximately normal. Construct a 95% confidence interval for the population mean mineral content. (Round the results to the nearest tenth if applicable.)