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m 2 MAT 1- x -t webassign.net/web/Student/Assignment Responses/last dep 15941748 20 points BBBasicStat7 76020 suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that Anterests us. (a) Suppose n 30 and p 0.30. Can we approximate p by a normal distribution? Why? (use 2 decimar places.) ESenech-M, BESeiect be approximated by a normal random variable because -Select What are the values of and oi? (Use 3 decimal places.) (b) suppose n 25 and p o.15, safely approximate a normal distribution? wby why not? Seieci M.PE select be approximate by a normal random variable because (c) Suppose n sy and p o 31. Can we aperoximate A by a normal distrouton? wny? Use decimal places.) normal random variable What are the values ot and se dec places

Explanation / Answer

1) a) n = 30 , p = .30 , q = 1- p = 1- 0.30 = .70

np = 30 * .30 = 9

nq = 30 * .70 = 21

mean = np = 9

std. deviation = npq = 6.3

c) n =53 , p = 0.33 , q = .67

np = 53 * .33 = 17.49

nq = 53 * .67 = 35.51

mean = np = 17.49

std. deviation = npq = 53 * .67 * .33 = 11.718

2)

mean =9.5 , std. deviation = 2.3

a) x = 10

By normal distribution formula,

z = ( x - mean) / s

= (10 - 9.5) / 2.3

= 0.217

Now, we need to find p( z <.217)

P(x <10) = p(z <.217) = .586

b) x = 5

By normal distribution formula,

z = ( x - mean) / s

= (5 - 9.5) / 2.3

= -1.95

Now, we need to find p( z > - 1.95)

P(x >5) = p(z > -1.95) = .9748

c) between 8 and 15

By normal distribution formula,

z = ( ( x - mean) / s < z <  ( x - mean) / s)

= ((8 - 9.5) / 2.3 < z < (15 - 9.5) / 2.3)

= (-0.65 < z < 2.39)

now, we need to find p( -0.65 < z < 2.39)

P( 8 < x <15) = p( -0.65 < z < 2.39) = .7345

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