Question

The proportion of business majors among college graduates in the US is 16%. Assume you select 27 college graduates at random. 1st write the formula with factorials for the probability of getting 8 business majors:

1. Compute the probability in the previous question:

   		0.0293
   		0.9794
   		0.0347
   		0.0127
   		0.9873

2. Compute the probability of getting 8 or less and 9 or more business majors:

   		0.9653, 0.0347
   		0.9794, 0.0206
   		0.0347, 0.9653
   		0.9873, 0.0127
   		0.0206, 0.9794

3. What are the expected value and standard deviation of this distribution? Which formulas to use?

   		4.32, 1.905 F6d, F6e
   		7.29, 1.776 F6d, F6e
   		7.29, 1.776 F6a, F6b
   		1.776, 7.29 F6a, F6b
   		1.905, 4.32 F6d, F6e

4. What is the range of usual values (in terms of a possible number of uccesses)?

   		3, 4, 5, 6, 7, 8, 9
   		2, 3, 4, 5, 6, 7, 8, 9
   		4, 5, 6, 7, 8, 9, 10
   		1, 2, 3, 4, 5, 6, 7, 8
   		0, 1, 2, 3, 4, 5, 6, 7

Answer 1

Compute the probability in the previous question:

n = 27, p = 0.16, or q = 1-p = 1 - 0.16 = 0.84

As per binomial distribution formula p(r) = ⁿC_r * p^r * q^n-r

p(8) = ²⁷C₈ (0.16)⁸ (0.84)¹⁹

p (8) = (27!/8! 19!) (0.16)⁸ (0.84)¹⁹

p(8) = 0.0347

Hence, Alternative third is correct.

2. Compute the probability of getting 8 or less and 9 or more business majors:

p(8 or less or 9 or more) :

p(8 or less) = p(0) + p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) + p(8)

p(8 or less) = (²⁷C₀ * 0.16⁰ * 0.84²⁷) + (²⁷C₁ * 0.16¹ * 0.84²⁶) + (²⁷C₂ * 0.16² * 0.84²⁵) + (²⁷C₃ * 0.16³ * 0.84²⁴) + (²⁷C₄ * 0.16⁴ * 0.84²³) + (²⁷C₅ * 0.16⁵ * 0.84²²) + (²⁷C₆ * 0.16⁶ * 0.84²¹) + (²⁷C₇ * 0.16⁷ * 0.84²⁰) + (²⁷C₈ * 0.16⁸ * 0.84¹⁹)

p(8 or less) = 0.009 + 0.0464 + 0.115 + 0.1825 + 0.2085 + 0.1827 + 0.1276 + 0.0729 + 0.0347

p(8 or less) = 0.9794

p(9 or more) = 1 - p(8 or less)

p(9 or more) = 1 - 0.9794

p(9 or more) = 0.0206

Thus, Alternative Second is correct.

3. What are the expected value and standard deviation of this distribution? Which formulas to use?

Expected Value = np = 27 × 0.16 = 4.32

Standard deviation =

Standard deviation = 1.905

Formula Used :

Expected value = F6d

Standard deviation = F6e

1.905, 4.32 F6d, F6e

Thus, Last Alternative is correcct.

4. What is the range of usual values (in terms of a possible number of uccesses)?

p(0) = 0.0090 < 0.05

p(1) = 0.0464 < 0.05

p(2) = 0.1149 > 0.05

p(7) = 0.0729 < 0.05

p(8) = 0.0347 < 0.05

p(9) = 0.0139 < 0.05

Thus, Range of usual Value is 1, 2, 3, 4, 5, 6, 7, 8

Hence, Alternative fourth is correct.