Most college-bound students take either the SAT(Scholastic Assessment Test) or the ACT (which originally stood for American coll
1 answer:
Answer:
Luis would need to have a SAT score of 574.
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.
Nicole's z-score:
ACT scores have a mean of about 21 with a standard deviation of about 5, which means that
Nicole gets a score of 24, which means that . Her z-score is:
What score would Luis have to have on the SAT to have the same standardized score(z-score) as Nicole's standardized score on the ACT?
Luis would have to get a score with a z-score of 0.6, that is, X when Z = 0.6.
SAT scores have a mean of about 508 with a standard deviation of about 110, which means that .
The score is:
Luis would need to have a SAT score of 574.
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Answer:
The 99% confidence interval for the proportion of ASD in Arizona is (0.014, 0.018).
Step-by-step explanation:
The information provided is as follows:
The sample proportion is:
The critical value of <em>z</em> for 99% confidence level is, <em>z</em> = 2.56.
Compute the 99% confidence interval for the proportion of ASD in Arizona as follows:
Thus, the 99% confidence interval for the proportion of ASD in Arizona is (0.014, 0.018).
Answer:
(a) The probability that the port handles less than 5 million tons of cargo per week is 0.7291.
(b) The probability that the port handles 3 or more million tons of cargo per week is 0.9664.
(c) The probability that the port handles between 3 million and 4 million tons of cargo per week is 0.2373.
(d) The number of tons of cargo per week that will require the port to extend its operating hours is 5.35 million tons.
Step-by-step explanation:
Let <em>X</em> = amount of cargo the port handles per week.
The random variable <em>X</em> is Normally distributed with parameters,
<em>μ</em> = 4.5 million
<em>σ</em> = 0.82 million
(a)
Compute the probability that the port handles less than 5 million tons of cargo per week as follows:
*Use a <em>z</em>-table for the probability.
Thus, the probability that the port handles less than 5 million tons of cargo per week is 0.7291.
(b)
Compute the probability that the port handles 3 or more million tons of cargo per week as follows:
*Use a <em>z</em>-table for the probability.
Thus, the probability that the port handles 3 or more million tons of cargo per week is 0.9664.
(c)
Compute the probability that the port handles between 3 million and 4 million tons of cargo per week as follows:
Thus, the probability that the port handles between 3 million and 4 million tons of cargo per week is 0.2373.
(d)
It is provided that P (X < x) = 0.85.
Then, P (Z < z) = 0.85.
The value of <em>z</em> for this probability is:
<em>z</em> = 1.04.
Compute the value of <em>x</em> as follows:
Thus, the number of tons of cargo per week that will require the port to extend its operating hours is 5.35 million tons.