What is log base 5(4*7 )+log base 5 of 2 written as a single log?
2 answers:
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Answer:
Proportion of the students recalled more than 15 names is 91.77%.
Step-by-step explanation:
We are given that a researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playing the name game, the students were asked to recall as many first names of fellow students as possible.
The mean number of names recalled was 19.41 with a standard deviation of 3.17.
<em>Let X = number of names recalled</em>
SO, X ~ N()
The z-score probability distribution is given by ;
Z = ~ N(0,1)
where, = mean number of names recalled = 19.41
= standard deviation = 3.17
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.
Now, proportion of the students recalled more than 15 names is given by = P(X > 15 names)
P(X > 15) = P( >
) = P(Z > -1.39)
= P(Z < 1.39) = 0.9177 {using z table}
<em>Therefore, proportion of the students recalled more than 15 names is </em><em>91.77%.</em>
Answer:
C. The distribution for town A is symmetric, but the distribution for
town B is negatively skewed.
Step-by-step Explanation:
From the box plots attached in the diagram below, which shows data of low temperatures for town A and town B for some days, we can compare the shapes of the box plot by visually analysing both box plots and how the data for each town is distributed.
=> For town A, the shape of the box plot is symmetric because both quartiles seem equal, and the median also divides the rectangular box into two equal halves. Both whiskers also appear to be of equal lengths.
The box plot for Town A takes a symmetric shape, and this shows a typical normal distribution of data.
=> On the other hand, Town B data distribution is different. The median seem close to the top half of the box and does not divide the box into equal halves. This shows the distribution is skewed. Since the whisker is shorter from the upper end of the box to the left side, we can infer that the distribution for Town B is skewed to the left, and it is negatively skewed.
=> The right comparison of the shapes of the box plots is "C. The distribution for town A is symmetric, but the distribution for town B is negatively skewed."
