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2 years ago

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A test was done to look at the maximum speed of mopeds. They found that the maximum speed had a normal distribution with a mean

of 29 mph and standard deviation of 1.4 mph. Consider randomly selecting a moped.
a. What is the probability that the maximum speed is at most 31 mph?
b. What is the probability that the maximum speed is at least 29.8 mph?
c. What is the probability that maximum speed differs from the mean value by at most 1.5 standard deviations?
d. Suppose we select 49 mopeds, what is the probability that the sample mean will be at least 29.5 mph?

1 answer:

Ksenya-84 [330]2 years ago

8 0

Answer:

C

Step-by-step explanation:

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Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below: Team X: 11, 3, 0,

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Answer:

C. Team Y’s scores have a lower mean value.

Step-by-step explanation:

We are given that Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below:

Team X: 11, 3, 0, 0, 2, 0, 6, 4

Team Y: 4, 2, 0, 3, 2, 1, 6, 4

Firstly, we will calculate the mean, median, range and inter-quartile range for Team X;

Mean of Team X data is given by the following formula;

Mean, $\bar X$ = $\frac{\sum X}{n}$

= $\frac{11+ 3+ 0+ 0+ 2+ 0+ 6+ 4}{8}$ = $\frac{26}{8}$ = 3.25

So, the mean of Team X's scores is 3.25.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team X: 0, 0, 0, 2, 3, 4, 6, 11

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2} )^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

Here, the number of observations is even, i.e. n = 8.

So, Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.} }{2}$

= $\frac{2+3}{2}$ = 2.5

So, the median of Team X's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

Range = Highest value - Lowest value

= 11 - 0 = 11

So, the range of Team X's score is 11.

Now, the inter-quartile range of the data is given by;

Inter-quartile range = $Q_3-Q_1$

$Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}$

= $(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(2.25 )^{th} \text{ obs.}$

$Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]$

= 0 + 0.25[0 - 0] = 0

$Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}$

= $3(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(6.75 )^{th} \text{ obs.}$

$Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]$

= 4 + 0.75[6 - 4] = 5.5

So, the inter-quartile range of Team X's score is (5.5 - 0) = 5.5.

<u>Now, we will calculate the mean, median, range and inter-quartile range for Team Y;</u>

Mean of Team Y data is given by the following formula;

Mean, $\bar Y$ = $\frac{\sum Y}{n}$

= $\frac{4+ 2+ 0+ 3+ 2+ 1+ 6+ 4}{8}$ = $\frac{22}{8}$ = 2.75

So, the mean of Team Y's scores is 2.75.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team Y: 0, 1, 2, 2, 3, 4, 4, 6

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2} )^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

Here, the number of observations is even, i.e. n = 8.

So, Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.} }{2}$

= $\frac{2+3}{2}$ = 2.5

So, the median of Team Y's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

Range = Highest value - Lowest value

= 6 - 0 = 6

So, the range of Team Y's score is 6.

Now, the inter-quartile range of the data is given by;

Inter-quartile range = $Q_3-Q_1$

$Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}$

= $(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(2.25 )^{th} \text{ obs.}$

$Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]$

= 1 + 0.25[2 - 1] = 1.25

$Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}$

= $3(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(6.75 )^{th} \text{ obs.}$

$Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]$

= 4 + 0.75[4 - 4] = 4

So, the inter-quartile range of Team Y's score is (4 - 1.25) = 2.75.

Hence, the correct statement is:

C. Team Y’s scores have a lower mean value.

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