You are forming words in which the first letter is either P, B, or M; the second letter is either A or E; and the third letter is either N or T. If each possible word is written on a slip of paper and you choose one slip at random, what is the probability that the second letter is an A?

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Ответ:

davidebner2248

03.05.2022

Mathematics

Number of People Selected for this Hypothesis = 84

Mean =M = 26.9 cm

Standard Deviation = S= 2.1 cm

$Z_{Score}=\frac{X_{score}-M}{S}$

$Z=\frac{84-26.9}{2.1}\\\\ Z=\frac{57.1}{2.1}\\\\ Z=27.190$

$Z_{95 percent}=1.96$

Confidence Interval for this Data will be

27.190 -1.96 ≤ Z ≤ 27.190 +1.96

25.230 ≤ Z ≤ 29.15

Option C: 22.784 to 31.016.

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