Based on data from a college, scores on a certain test are normally distributed with a mean of 1530 and a standard deviation of
1 answer:
Answer:
0.73% of the scores are greater than 2317.
14.46% of the scores are less than 1190.
38.23% of the scores are between 1351 and 1673.
Step-by-step explanation:
Problems of normally distributed samples can be 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 percentage of scores greater than 2317.
This is 1 subtracted by the pvalue of Z when X = 2317. So:
has a pvalue of 0.9927.
So 1-0.9927 = 0.0073 = 0.73% of the scores are greater than 2317.
Find the percentage of scores less than 1190.
This is the pvalue of Z when X = 1190. So:
has a pvalue of 0.1446.
So 14.46% of the scores are less than 1190.
Find the percentage of scores between 1351 and 1673.
This is the pvalue of Z when X = 1673 subtracted by the pvalue of Z when X = 1351. So
X = 1673
has a pvalue of 0.67
X = 1351
has a pvalue of 0.2877
So 0.67-0.2877 = 0.3823 = 38.23% of the scores are between 1351 and 1673.
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Answer:
Step-by-step explanation:
Let x represent the random variable representing the scores in the exam. Given that the scores are normally distributed with a mean of 40 and a standard deviation of 5, the diagram representing the curve and the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations is shown in the attached photo
1 standard deviation = 5
2 standard deviations = 2 × 5 = 10
3 standard deviations = 3 × 5 = 15
1 standard deviation from the mean lies between (40 - 5) and (40 + 5)
2 standard deviations from the mean lies between (40 - 10) and (40 + 10)
3 standard deviations from the mean lies between (40 - 15) and (40 + 15)
b) We would apply the probability for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = sample mean
µ = population mean
σ = standard deviation
From the information given,
µ = 40
σ = 5
the probability that a randomly selected score will be greater than 50 is expressed as
P(x > 50) = 1 - P( ≤ x 50)
For x = 50,
z = (50 - 40)/5 = 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.98
P(x > 50) = 1 - 0.98 = 0.02
Answer:
x-coordinates of relative extrema =
x-coordinates of the inflexion points are 0, 1
Step-by-step explanation:
Differentiate with respect to x
Differentiate f'(x) with respect to x
At x = ,
We know that if then x = a is a point of minima.
So, is a point of minima.
For inflexion points:
Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.
So, x-coordinates of the inflexion points are 0, 1
Answer
a) C
b) B
c) C, E
d) 15.5
e) 0.7
f) 67.7
Step-by-step explanation
a) A linear equation can be written as where
is the slope and
is the intercept on the
-axis. The slope describes how
changes when
changes. A negative value means the it is a decreasing change. In our case,
is
and
is
. Thus, the slope of the equation is
. This means that the area of glacier is decreasing by
every year since 2000.
b) The -intercept (= 16.2
) represents the area covered when
. But
starting from year 2000. This intercept then represents the area covered by glacier in the year 2000. Therefore, the area covered by glacier in 2000 was 16.2
.
c) means the area covered after year 2000. Setting it to 12 means the area covered after
years is 12
).
is therefore the number of years since 2000 that the total area covered will be 12
).
d) to 1 decimal place.
e) The term involving represents the disappearing area. In 12 years, the disappeared area is
to 1 decimal place.
f)
years to 1 decimal place.