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A professor has kept records on grades that students have earned in his class. If he wants to examine the percentage of students

AleksandrR [38]

Answer:

Pie chart

Step-by-step explanation:

Given that a professor has kept records on grades that students have earned in his class

He wants to examine the percentage of students earning the grades A, B, C, D, and F during the most recent term,

We have to find which kind of plot could he make

Box plot gives comparison for interval of scores this is to some extent helpful to find out the percentage of students getting a range of scores

Timeplot is irrelevant here. Dot plot will not be of much use.

Histogram is to some extent useful as box plot

But the best one is pie chart giving easy visual comparison for various grades A, B,C,D,E and F

8 0

3 years ago

How to divided 4,44 by 3

marysya [2.9K]

444 divided by 3 equals 148

8 0

3 years ago

Read 2 more answers

The mean height of women in a country (ages 20-29) is 64 4 inches A random sample of 50 women in this age group is selected What

Simora [160]

Answer:

0.0721 = 7.21% probability that the mean height for the sample is greater than 65 inches.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .