Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation, also called standard error of the mean [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
(a) Is the distribution of scores on this nal exam symmetric, right skewed, or left skewed?
Mean = 70, median = 74. So the distribution is left-skewed.
(b) Would you expect most students to have scored above or below 70 points?
70 is below the median, which is 74.
50% score above the median, and 50% below. So 50% score above 74.
This means that we should expect most students to have scored above 70.
(c) Can we calculate the probability that a randomly chosen student scored above 75 using the normal distribution?
The scores are skewed, so we cannot calculate any probability for a single student.
(d) What is the probability that the average score for a random sample of 40 students is above 75?
Now we can apply the central limit theorem.
[tex]\mu = 70, \sigma = 10, n = 40, s = \frac{10}{\sqrt{40}} = 1.58[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 75. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central limit theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{75 - 70}{1.58}[/tex]
[tex]Z = 3.16[/tex]
[tex]Z = 3.16[/tex] has a pvalue of 0.9992
1 - 0.9992 = 0.0008
0.08% probability that the average score for a random sample of 40 students is above 75
(e) How would cutting the sample size in half aect the standard error of the mean?
n = 40
[tex]s = \frac{10}{\sqrt{40}} = 1.58[/tex]
n = 20
[tex]s = \frac{10}{\sqrt{20}} = 2.24[/tex]
If the sample size is cut in half, the standard error of the mean would increase fro 1.58 to 2.24.
