F(x) = -22 – 92 = 9x Find f(-2)
1 answer:
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Answer:
[0.9 months, 32.69 months]
Step-by-step explanation:
The mean is
The standard deviation is
Now, we have to find two values a and b such that the area under the Normal curve with mean 16.8 and standard deviation 8.1092 between a and b equals <em>95% = 0.95 </em>
Using a spreadsheet we find these values are
a = 0.906
b = 32.694
<h3>(See picture) </h3>and our 95% confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program rounded to two decimal places is
[0.9 months, 32.69 months]
Answer:
Vanessa records a total of <u>11</u> quiz scores
Moe records a total of <u>10</u> quiz scores
According to the data, Vanessa's mean score is 80
According to the data, Vanessa's median score is 85
The range of Moe's quiz score is 45
The interquartile range of Moe's quiz score is 16.25
Step-by-step explanation:
Vanessa's score is given in the dot plot
Moe's scores are; 75, 80, 90, 90, 95, 55, 100, 95, 85, 85
1) The number of quiz scores Vanessa records is given by the number of dots on the dot plot, n = 11 dots ≡ 11 quiz scores
2) The number of quiz scores Moe records is given by the number of data points, n = 10 data points ≡ 10 quiz scores
3) Vanessa's mean score = ∑x/n
∑x/n = (60 + 2× 65 + 75 + 4 × 85 + 2 × 90 +95)/11 = 80
Vanessa's mean score = 80
4) Vanessa's median score is the (11 + 1)/2 th score = Her 6th score = 85
Vanessa's median score = 85
5) The range of Moe's score = Her largest score - Her least score = 100 - 55 = 45
The range of Moe's score = 45
6) Sorting Moe's scores gives;
75, 80, 90, 90, 95, 55, 100, 95, 85, 85
55, 75, 80, 85, 85, 90, 90, 95, 95, 100
The first quartile, Q₁ = 11/4th score = 2.75th score = 75 + 0.75 × 5 = 78.75
The 3rd quartile, Q₃ = The 3/4×11 score = 8.25th score = 95 + 0.25 ×0 = 95
The interquartile range, I = Q₃ - Q₁ = 95 - 78.75 = 16.25
The interquartile range of Moe's quiz score = 16.25.