Question

In 2016, the mean salary for all Ontario public employees in the School Board sector with salaries of at least $100,000 was 114.35 (in thousands of dollars), and the standard deviation was 15.67 (also in thousands of dollars). Let ??Xi be the salary of one randomly selected employee (denoted i), and let ?¯X¯ be the sample mean for a sample of 500 employees with salaries of at least $100,000 from the School Board sector.

(i) Which is larger: ?(??>116)P(Xi>116) or ?(?¯>116)?P(X¯>116)? Explain. (ii) Compute any probabilities possible with the given information.

(b) [6 pts] Suppose we know that ?(??>120)=?P(Xi>120)=a, where ?a is a constant. We define a new random variable ??Yi for each i, where ??Yi is equal to 1 if ??>120Xi>120 and is equal to 0 otherwise (??Xi is defined in part (a)). Also let ?=∑200?=1??W=∑i=1200Yi, where we assume that all the ??Yi’s are independent and have the same distribution.

What kind of distribution does ??Yi have (hint: it has a name)? What are the values of its parameter(s)? What kind of distribution does ?W have (it also has a name)? What are the values of its parameters? Explain in 1 sentence how you know ?W has this distribution.

(c) [6 pts] Suppose now that ?=0.3a=0.3. What is ?(?>70)P(W>70)? Use an appropriate approximation and correct for continuity. Give a clear interpretation (1 sentence).

Answer 1

Answer:

Given that,

In 2016, the mean salary for all Ontario public employees in the School Board sector with salaries of at least $100,000 was 114.35 (in thousands of dollars), and the standard deviation was 15.67 (also in thousands of dollars).

Let Xi be the salary of one randomly selected employee (denoted i), and let be the sample mean for a sample of 500 employees with salaries of at least $100,000 from the School Board sector.

(a).

(i). Which is larger: P(Xi>116) or P( >116). Explain,(ii). Compute any probabilities possible with the given information:

P(Xi > 116) = P(Z<(116-114.35)/15.67)

= P(Z> (1.65/15.67))

=P( Z > 0.1053)

=1-P(Z < 0.1053)

=1-0.5434

= 0.4562

Here n=500

=P( Z > 2.3545)

=1-P( Z < 2.3545)

=1-0.9906

=0.0094

i.e. P(Xi > 116) is larger.

(b).

Y has Bernoulli distribution with:

Probability of success = p = P(X>120)

= P(Z>(120-114.35)/15.67)

= P(Z>(5.65)/15.67)

=P( Z > 0.3606)

=1- P( Z < 0.3606)

=1-0.6406

= 0.3594

W has a binomial distribution with:

Probability of success = p = P(X>120)

= 0.3594

and n = 200

W satisfies all the conditions of binomial distribution like p remaining consistent among all trials, independent trials, etc.

(c).

Suppose now that a=0.3.

What is P(W>70):

P( W > 70)= P(70 W 200)

The population mean is computed as:

=200 0.3

=60

and the population standard deviation is computed as:

=6.4807

Therefore, we get that

=0.0526

So then, we conclude that P(70 W 200)=0.0526, which ends the calculation of the requested probability.