Question

. Among all the income tax forms filed in a certain year, the mean tax paid was $2000, and the standard deviation was $500. In addition, for 10% of the forms, the tax paid was greater than $3000. A random sample of 625 tax forms is drawn. a. What is the probability that the average tax paid on the sample forms is greater than $1980? b. What is the probability that more than 60 of the sampled forms have a tax of greate

Answer 1

Answer:

a) 0.8413

b) 0.6293

Step-by-step explanation:

Data provided:

Mean tax paid = $2000

Standard deviation = $500

Sample size, n = 625

n=625

Now,

a) P( average tax paid on the sample forms is greater than $1980)

⇒ P(X > 1980)

or

⇒ $P(\frac{(X-mean)}{\frac{s}{\sqrt n}})$

or

⇒ $P(\frac{(1980-2000)}{\frac{500}{\sqrt{625}}})$

or

⇒ P(Z > -1)

or

= 0.8413 (From standard normal table)

b) P(more than 60 of the sampled forms have a tax of greater than $3000)

given:  p = 10% = 0.1

Now, by using CLT

mean = n × p

= 625 × 0.1

= 62.5

Standard deviation, s =$\sqrt{n\times p\times(1-p)}$

=$\sqrt{625\times0.1\times(1-0.1)}$

= 7.5

Thus,

P(X > 60)

= $P(\frac{(X-mean)}{s} > \frac{(60-62.5)}{7.5})$

or

= P(Z > -0.33)

= 0.6293     (From standard normal table)