Answer:
The median and mode of the students grade is 70.
Most of the students scored between 40 and 100.
Step-by-step explanation:
From the provided information it can be seen that the mean of the distribution is, <em>μ</em> = 70 and the standard deviation is, <em>σ</em> = 15.
For a Normal distributed data the mean, median and mode are the same.
So, the median and mode of the students grade is 70.
The standard deviation of the data represents the spread of the observation, i.e. how dispersed the values are along the curve.
In statistics, the 68–95–99.7 rule, also recognized as the empirical rule, is a shortcut used to recall that 68.27%, 95.45% and 99.73% of the values of a Normally distributed data lie within one, two and three standard deviations of the mean, respectively.
Assuming that maximum marks of the exam is 100, it can be said that most of the students scored between 40 and 100.
Answer:
Hey there!
We can't compare two measurements without converting them to the same units. Thus, we use proportions to make all the values into the same unit.
, which converts to centimetres to millimetres.
12.5 cm=125 mm.
Now, we can compare the values of 125 mm and 140 mm.
Clearly, we see that 140 mm is greater than 125 mm.
Let me know if this helps :)
Answer:
35.2% probability that the sample mean will be 246 pages or more
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
What the probability that the sample mean will be 246 pages or more?
This is 1 subtracted by the pvalue of Z when X = 246. So
By the Central Limit Theorem
has a pvalue of 0.6480.
1 - 0.6480 = 0.3520
35.2% probability that the sample mean will be 246 pages or more