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4. In written English, the letter .e, is 10% of all individual written letters. (a) A linguist selects a random sample of 298
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Answer #1

A.

n = 298

\hat{p} = 0.1

center = 0.1

Std. Deviation = 0.0174 (VP(1-p) /n

When np.(i.е., X) 5 and n(1-p) (i.e., n-X) 25, sampling distribution of p can be approximated bya normal distribution with me

Since,

np = 29.8> 5 and

n(1-p) = 268.2 > 5

Thus assumptions to consider sampling distribution of \hat{p} approximately normal are fulfilled.

2.

P(\hat{p}>0.12) = 0.1252

0.048 0.065 0.083 0.1 0117 0.135 0.152

Let z be the standard normal variable. so,

Z = (\hat{p}-p)/\sigma _{\hat{p}}

P(\hat{p}>0.12) = p(Z>(0.12-0.1)/0.0174)

= P(Z>1.1494)

= 0.1252

Please upvote if you have liked my answer, would be of great help. Thank you.

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