Question

The scores of all applicants taking an aptitude test required by a law school have a normal distribution with a mean of 420 and a standard deviation of 100. A random sample of 25 scores is taken.

e. The probability is 0.05 that the sample standard deviation of the scores is higher than what number?

f. The probability is 0.05 that the sample standard deviation of the scores is lower than what number?

g. If a sample of 50 test scores had been taken, would the probability of a sample mean score higher than 450 be smaller than, or larger than, or the same as the correct answer to part a? it is not necessary to do the detailed calculations here. Sketch a graph to illustrate your reasoning

a. Find the probability that the sample mean score is higher than 450. (I believe the answer is 0.068)

Answer 1

a)

The scores of students taking aptitude test at a particular institute, X, are normally distributed with mean 420 and standard deviation of 100. We are considering a random sample of size 25 from above population. Here we have: u = 420 o=100 n = 25

$\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{25}}=20$

a. The calculation of the probability is as shown below: z X-u P(X > 450)=1-pZ34 = 1-plz 450-420 100 25 =1-p(Z 31.5) The calculation ofl-p(Z 31.5) in Excel is: fx =1-NORMSDIST(1.5) DE F 0.0668 Therefore, the probability that the sample mean score is higher than 450 is 0.0668.

e)

e) Consider the sample standard devotion. We know that the mean of sample standard deviation is E(52) = 100and the variance is Var(s) = 2*1.00* = 8,333,333.33. Thus, x = 24.5 has chi-square distribution with 24 degrees of freedoms. We are interested in finding the value above which 5% of the sample standard deviations of scores lie. Thus we need to find the value of s2 such that Pr(S? > s?)=0.05. Pr(s? >s?) = 0.05 Pr(SS82)=0.95 From the chi square distribution table at 24 degrees of freedom we know that: Pr(x: 536.42)=0.95 So, we have: 24s 1002 = 36.42 s= 3642X100 36.42x1002 = /15,175 = 123.19 Thus, the value, above which 5% of the sample standard deviations of scores lie, is 123.19.

f)

f) We are interested in finding the value below which 5% of the sample standard deviations of scores lie. Thus we need to find the value of s? such that Presss?)=0.05. Pr(s? <s?) = 0.05 Po( z* = 2a.-0.05 From the chi square distribution table at 24 degrees of freedom we know that: Pr(x24 513.85)=0.05

So, we have: 2452 1002=13.85 13.85 x 1002 SE 24 = 5.770.83 = 75.97 Thus, the value, below which 5% of the sample standard deviations of scores lie, is 75.97.

g)

g) Consider that instead of sample of size 25 we take a sample of size 50. It follows that the standard deviation of the sample mean will reduce with increase in sample size. Thus, the probability that the sample mean score is above 450 will be lesser when sample size is 50 as compared to the probability when the sample size is 25. This can be further clarified using the figure below. Sampling Distribution of Sample Mean 0.0301 0.025 0.020 Density 0.015 0.010 0.005 0.0004 350 375 400 425 450 475 Sample size 25 Sample size 50 –

We can observe that the probability under the red curve is smaller than the probability under black curve

Please upvote if this helps