Well you have to have the common at 50 because that it so multiply23/25 by 2 and 9/10 by 5 so 45/50 and 46/50
Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.
<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.
The mean and the standard deviation are given, respectively, by:
The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:
X = 590:
Z = 0.76
Z = 0.76 has a p-value of 0.7764.
X = 400:
Z = -0.89
Z = -0.89 has a p-value of 0.1867.
0.7764 - 0.1867 = 0.5897 = 58.97%.
58.97% of students would be expected to score between 400 and 590.
More can be learned about the normal distribution at brainly.com/question/27643290
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Answer:
x = 80.96
Step-by-step explanation:
9x * 3x +1 = 2187
27x +1 = 2187
27x = 2187-1
27x = 2186
x = 2186/27 = 80.96(2dp)
Answer:
missing side length = 9
Step-by-step explanation:
a = 6
b = 7
c = missing side
Answer:
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. 16)n = 182, x = 135; 95 percent
✓ 16)n = 182, x = 135; 95 percent sample proportion: p-hat = 135/182 = 0.74 E = 1.96*sqrt[0.74*0.26/182] = 0.0637 95% CI: 0.74-0.0637 < p < …