Question

73% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places.

If 32 owned dogs are randomly selected, find the probability that

a. Exactly 21 of them are spayed or neutered.

b. At most 22 of them are spayed or neutered.

c. At least 24 of them are spayed or neutered.

d. Between 20 and 25 (including 20 and 25) of them are spayed or neutered.

The highway mileage (mpg) for a sample of 8 different models of a car company can be found below. Find the mean, median, mode, and standard deviation. Round to one decimal place as needed.

19, 22, 26, 28, 30, 32, 34, 34

Mean =

Median =

Mode =

Standard Deviation =

Answer 1

1)

The population proportion of success is p = 0.73,

also, 1 - p = 1 - 0.73 = 0.27

and the sample size is n= 32.

a)

b. At most 22 of them are spayed or neutered.

Pr[ X 22] = 1 - Pr(X≥23)

We need to compute Pr(X≥23)

Therefore, we get that

This implies that

which means that the probability we are looking for is Pr(X≥23)=0.6439.

Pr[ X 22] = 1 - Pr(X≥23)

= 1 - 0.6439

= 0.3561

c)

We need to compute Pr(X≥24)

This implies that

which means that the probability we are looking for is

Pr(X≥24)=0.4902.

d) d. Between 20 and 25 (including 20 and 25) of them are spayed or neutered.

Therefore, we get that

This implies that

which means that the probability we are looking for is

Pr(20≤X≤25)=0.7343.

2)

The sample size is n = 8.

The provided sample data along with the data required to compute the sample mean Xˉ and sample variance s^2 are shown in the table below:

	mpg	mpg2
	19	361
	22	484
	26	676
	28	784
	30	900
	32	1024
	34	1156
	34	1156
Sum =	225	6541

The sample mean Xˉ is computed as follows:

Also, the sample variance s^2 is

Therefore, the sample sandartd deviation ss is

Median = the point which divides the data into two equal parts

here n = 8 so we take average of two middle values.

median =

median =

Mode = data which occur most number of the times

in our sample 34 repeated twice .

hence

mode = 34

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