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The round-off errors when measuring the distance that a long jumper has jumped are uniformly distributed between 0 and 5.4 mm. Round values to 4 decimal places when possible.

a. The mean of this distribution is 2.7.

b. The standard deviation is 1.5583.

c. The probability that the round-off error for a jumper's distance is exactly 0.4 is [tex]P(x = 0.4) = \square[/tex]

d. The probability that the round-off error for the distance that a long jumper has jumped is between 1.8 and 5.1 mm is [tex]P(1.8 \ \textless \ x \ \textless \ 5.1) = 0.6111 \square[/tex]

e. The probability that the jump's round-off error is greater than 1.28 is [tex]P(x \ \textgreater \ 1.28) = 0.7630[/tex]

f. [tex]P(x \ \textgreater \ 3.71 \, | \, x \ \textgreater \ 0.9) = \square[/tex]

g. Find the 55th percentile. [tex]\square[/tex]

h. Find the maximum for the lower quartile. [tex]\square[/tex]

Sagot :

GinnyAnswer
To solve the given problem about the uniform distribution of round off errors when measuring the distance that a long jumper has jumped, let's break it down step by step. Please note that values are rounded to four decimal places as required.

### a. Mean of the Distribution
For a uniform distribution between 0 and 5.4 mm, the mean is calculated as:
[tex]\[ \text{Mean} = \frac{\text{Upper Bound} + \text{Lower Bound}}{2} = 2.7000 \][/tex]

### b. Standard Deviation of the Distribution
The standard deviation for a uniform distribution is calculated using the formula:
[tex]\[ \text{Standard Deviation} = \sqrt{\frac{(\text{Upper Bound} - \text{Lower Bound})^2}{12}} = 1.5583 \][/tex]

### c. Probability of Exactly 0.4 mm Error
For a continuous uniform distribution, the probability of any specific value (like exactly 0.4 mm) is 0:
[tex]\[ P(x=0.4) = 0 \][/tex]

### d. Probability Between 1.8 mm and 5.1 mm
The probability that the round off error is between 1.8 mm and 5.1 mm can be found using:
[tex]\[ P(1.8 < x < 5.1) = \frac{5.1 - 1.8}{5.4 - 0} = 0.6111 \][/tex]

### e. Probability Greater Than 1.28 mm
To find the probability that the error is greater than 1.28 mm, use:
[tex]\[ P(x > 1.28) = \frac{5.4 - 1.28}{5.4 - 0} = 0.7630 \][/tex]

### f. Probability Greater Than 3.71 mm or 0.9 mm
Since 3.71 mm is greater than 0.9 mm, the probability [tex]\(P(x > 0.9)\)[/tex] encompasses [tex]\(P(x > 3.71)\)[/tex]:
[tex]\[ P(x > 0.9) = \frac{5.4 - 0.9}{5.4 - 0} = 0.8333 \][/tex]

### g. The 55th Percentile
To find the 55th percentile:
[tex]\[ \text{55th Percentile} = 0 + 0.55 \times (5.4 - 0) = 2.9700 \][/tex]

### h. Maximum for the Lower Quartile (25th Percentile)
The maximum value for the lower quartile is positionally at 25% of the data:
[tex]\[ \text{Lower Quartile (25th Percentile)} = 0.25 \times (5.4 - 0) = 1.3500 \][/tex]

Thus, summarizing the results:

- c. [tex]\( P(x=0.4) = 0 \)[/tex]
- f. [tex]\( P(x > 3.71 \text{ or } x > 0.9) = 0.8333 \)[/tex]
- g. The 55th percentile = 2.9700
- h. The maximum for the lower quartile = 1.3500
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