Question

Suppose we have a binomial distribution with *n* trials
and probability of success *p*. The random variable
*r* is the number of successes in the *n* trials, and
the random variable representing the proportion of successes is
*p̂* = *r*/*n*.

(a) *n* = 44; *p* = 0.53; Compute *P*(0.30
≤ *p̂* ≤ 0.45). (Round your answer to four decimal
places.)

(b) *n* = 36; *p* = 0.29; Compute the probability
that *p̂* will exceed 0.35. (Round your answer to four
decimal places.)

(c) *n* = 41; *p* = 0.09; Can we approximate
*p̂* by a normal distribution? Explain.

---Select--- Yes No , *p̂* ---Select--- can
cannot be approximated by a normal random variable because

Answer #1

(a) Given : *n* = 44, *p* = 0.53

Now ,

; From standard normal distribution table

(b) Given : n=36 , p=0.29

Now ,

; From standard normal distribution table

(c) Given : n=41 ,p=0.09

Here , np=41*0.09=3.69<5

Therefore , we cannot be approximated by a normal random variable because np<5

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p̂ ---Select--- be approximated by a normal random variable because
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np =
nq =
---Select--- Yes No
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np =
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us.
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nq =
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a. a Hypergeometric distribution with N =
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b. a Poisson distribution with mean 4.
c. an exponential distribution with mean 4.
d. another binomial distribution with n =
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e.
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us.
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