A bunch of bananas cost $4, a pineapple cost $2 more than a bunch of bananas, and 4 apples cost one-half of a pineapple.

MA_775_DIABLO [31]

Yes, an additional 4% of Americans now support the President's policy on Afghanistan.

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3 years ago

Each year about 1500 students take the introductory statistics course at a large university. This year scores on the nal exam ar

nikitadnepr [17]

Answer:

a) Left-skewed

b) We should expect most students to have scored above 70.

c) The scores are skewed, so we cannot calculate any probability for a single student.

d) 0.08% probability that the average score for a random sample of 40 students is above 75

e) If the sample size is cut in half, the standard error of the mean would increase fro 1.58 to 2.24.

Step-by-step explanation:

To solve this question, we need to understand skewness,the normal probability distribution and the central limit theorem.

Skewness:

To undertand skewness, it is important to understand the concept of the median.

The median separates the upper half from the lower half of a set. So 50% of the values in a data set lie at or below the median, and 50% lie at or above the median.

If the median is larger than the mean, the distribution is left-skewed.

If the mean is larger than the median, the distribution is right skewed.

Normal probabilty distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation, also called standard error of the mean $s = \frac{\sigma}{\sqrt{n}}$