Question

Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.

(a) Suppose n = 43 and p = 0.33. Can we approximate p̂ by a normal distribution? Why? (Use 2 decimal places.)

np =

nq =

---Select--- Yes No , p̂  ---Select--- cannot can be approximated by a normal random variable because  ---Select--- nq exceeds np does not exceed both np and nq exceed nq does not exceed np and nq do not exceed np exceeds  .

What are the values of μ_p̂ and σ_p̂? (Use 3 decimal places.)

μp̂ =

σp̂ =

(b) Suppose n = 25 and p = 0.15. Can we safely approximate p̂ by a normal distribution? Why or why not?
---Select--- Yes No , p̂  ---Select--- can cannot be approximated by a normal random variable because  ---Select--- both np and nq exceed np exceeds nq does not exceed np does not exceed nq exceeds np and nq do not exceed  .

(c) Suppose n = 65 and p = 0.24. Can we approximate p̂ by a normal distribution? Why? (Use 2 decimal places.)

np =

nq =

---Select--- Yes No , p̂  ---Select--- can cannot be approximated by a normal random variable because  ---Select--- np does not exceed both np and nq exceed np exceeds nq does not exceed nq exceeds np and nq do not exceed  .

What are the values of μ_p̂ and σ_p̂? (Use 3 decimal places.)

μp̂ =

σp̂ =

Answer 1

a)

n p = 43 * 0.33 = 14.19

n q = 43 * ( 1 - 0.33) = 28.81

Yes, can be approximately normal random vaiable , because both np and nq exceeds 10

= p = 0.33

= sqrt ( p ( 1 - p) / n ) )

= sqrt ( 0.33 * ( 1 - 0.33) / 43)

= 0.072

b)

n p = 25 * 0.15 = 3.75

n q = 25 * ( 1 - 0.15) = 21.25

No, cannot be approximately normal random variable

because np does not exceeds and nq exceeds 10

c)

n p = 65 * 0.24 = 15.6

n q = 65 * ( 1 - 0.24) = 49.4

Yes, can be approximately normal because both np and nq exceeds 10

= p = 0.24

= sqrt ( p ( 1 - p) / n ) )

= sqrt ( 0.24 * ( 1 - 0.24) / 65)

= 0.053