You would distribute then combine like terms to get your answer.
Mischa and Ben's combined SAT score was 2290. Ben and Ashley's
1 answer:
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Answer:
The numerical limits for a B grade are 81 and 89, that is, a score between 81 and 89 gets a B grade.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Scores on the test are normally distributed with a mean of 79.7 and a standard deviation of 8.4.
This means that
B: Scores below the top 13% and above the bottom 56%
So between the 56th percentile and the 100 - 13 = 87th percentile.
56th percentile:
X when Z has a p-value of 0.56, so X when Z = 0.15. Then
87th percentile:
X when Z has a p-value of 0.87, so X when Z = 1.13.
The numerical limits for a B grade are 81 and 89, that is, a score between 81 and 89 gets a B grade.
Answer:
Step-by-step explanation:
Given that the housing market has recovered slowly from the economic crisis of 2008. Recently, in one large community, realtors randomly sampled 38 bids from potential buyers to estimate the average loss in home value.
s = sample std deviation = 3000
Sample mean = 9379
Sample size n = 38
df = 37
Std error of sample mean =
confidence interval 95% = Mean ± t critical * std error
=Mean ±1.687*486.66 = Mean ±821.003
=(8557.997, 10200.003)
a) If std deviation changes to 9000 instead of 3000, margin of error becomes 3 times
Hence 2463.008
b) The more the std deviation the more the width of confidence interval.