Using the normal distribution, it is found that there are 68 students with scores between 72 and 82.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean and standard deviation is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by:
The proportion of students with scores between 72 and 82 is the <u>p-value of Z when X = 82 subtracted by the p-value of Z when X = 72</u>.
X = 82:
Z = 1
Z = 1 has a p-value of 0.84.
X = 72:
Z = 0
Z = 0 has a p-value of 0.5.
0.84 - 0.5 = 0.34.
Out of 200 students, the number is given by:
0.34 x 200 = 68 students with scores between 72 and 82.
More can be learned about the normal distribution at brainly.com/question/24663213
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48 girls are there. Given that there is a ratio of boys to girls of 7:8. Then divide 42 boys by 7 boys which equals 6. After, multiply 6 by 8 girls equals 48 girls.
Answer:
(-162)/7 or -23 1/7 as mixed fraction
Step-by-step explanation:
Simplify the following:
(-36)/14 (-18) (-3)/6
Hint: | Express (-36)/14 (-18) (-3)/6 as a single fraction.
(-36)/14 (-18) (-3)/6 = (-36 (-18) (-3))/(14×6):
(-36 (-18) (-3))/(14×6)
Hint: | In (-36 (-18) (-3))/(14×6), divide -18 in the numerator by 6 in the denominator.
(-18)/6 = (6 (-3))/6 = -3:
(-36-3 (-3))/14
Hint: | In (-36 (-3) (-3))/14, the numbers -36 in the numerator and 14 in the denominator have gcd greater than one.
The gcd of -36 and 14 is 2, so (-36 (-3) (-3))/14 = ((2 (-18)) (-3) (-3))/(2×7) = 2/2×(-18 (-3) (-3))/7 = (-18 (-3) (-3))/7:
(-18 (-3) (-3))/7
Hint: | Multiply -18 and -3 together.
-18 (-3) = 54:
(54 (-3))/7
Hint: | Multiply 54 and -3 together.
54 (-3) = -162:
Answer: (-162)/7