A tabletop in the shape of a trapezoid has an area of 7,701 square centimeters. Its longer base measures 119 centimeters, and th
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Answer:
Grades between 62 and 64 result in a D grade.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
Find the numerical limits for a D grade.
D: Scores below the top 84% and above the bottom 10%
So below the 100-84 = 16th percentile and above the 10th percentile.
16th percentile:
This is the value of X when Z has a pvalue of 0.16. So X when Z = -0.995.
10th percentile:
This is the value of X when Z has a pvalue of 0.1. So X when Z = -1.28.
Grades between 62 and 64 result in a D grade.
Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: fitness of a line of E. coli grown on an acidic environment.
n= 6 E. coli lines
Recorded fitness for each line: 1.24, 1.22, 1.23, 1.24, 1.18, 1.09
The relative fitness of 1 indicates that both bacteria types are equally fit.
A relative fitness larger than 1 indicates that the acid-evolved line is more fit than the parental line kept at neutral pH when both are grown in acidic conditions.
Meaning that if the average fitness of the E. coli lines grown on an acidic environment is greater than 1 then they are better adjusted to live in acidic conditions, symbolically: μ > 1
The statistic hypotheses are:
H₀: μ ≤ 1
H₁: μ > 1
α: 0.05
Assuming that the variable has a normal distribution you have to apply a one-sample t-test:
X[bar]= 1.20
S= 0.06
The p-value for this test is 0.0002
Since the p-value= 0.0002 is less than α:0.05 the decision is to reject the null hypothesis.
Then at a 5% significance level, there is significant evidence to conclude that the bacteria evolved in acidic pH are better adapted to acidic conditions.
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