Question

You will need the same dataset used for problem 3 in homework 1 (the dataset obtained from the yahoo website with the company you selected). Use Excel to calculate the average and standard deviation of the close data column. Assume that these two numbers represent the population (parametric mean and population standard deviation, respectively, for the variable length (in cm) in a population printout of the data to your homework and write down the ticker code on it.
a. Calculate the probability of sampling at random a fish that is smaller in size than the value you would obtain by subtracting half the standard deviation from the average [x will be equal to: -(6/2)]
b. Calculate the probability of sampling at random a fish that is greater in size than the value you would obtain by adding half the standard deviation from the average [x = u + (0/2)]
c. Calculate the probability of sampling at random a fish that has a size between the two values [x = -(6/2), x=u +(6/2)) used in parts "a" and "b," respectively
d. Calculate the 25th and 75 percentiles of fish size for the population using the normal distribution table. e. Imagine that 5 individuals are sampled at random from this fish population. Calculate the probability that the average calculated will be less than the value: -(6/3)

Answer 1

Answer:

a) The probability of sampling at random a fish that is smaller in size than the value you would obtain by subtracting half the standard deviation from the average is 0.3085.

b) The probability of sampling at random a fish that is greater in size than the value you would obtain by adding half the standard deviation from the average is 0.3085.

c) The probability of sampling at random a fish that has a size between the two values is 0.383.

d) The 25th and 75 percentiles of fish size for the population using the normal distribution table is 5.69 and 5.87 respectively.

e) The probability that the average calculated will be less than the value is 0.3707.

Step-by-step explanation:

For the given data set mean $(\mu) = 5.75913$

Standard deviation $(\sigma) = 0.172229$

Variance $(\sigma2) = 0.0296$

Here we get is

a)

$P(x < \mu - \sigma/2) = p(x < 5.673) \\\\ = 0.3085$

b)

$P(x < \mu + \sigma/2) = p(x < 5.845) \\\\= 0.3085$

c)

$P(\mu - \sigma/2 < x < \mu + \sigma/2) = p(5.673 < x < 5.845) \\\\= 0.383$

d)

25th percentile:-

$= 25*[(n+1)/100]th term \\\\= 5.69$

75the percentile:-

$= 75*[(n+1)/100]th term\\\\ = 5.87$

e)  

$p(x < \mu - \sigma/3) = p(x < 5.7017) \\\\= 0.3707$