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elena-s [515]
2 years ago
12

Assume that 33.4% of people have sleepwalked. Assume that in a random sample of 1459 ​adults, 526 have sleepwalked.

Mathematics
1 answer:
lutik1710 [3]2 years ago
7 0

Answer:

Step-by-step explanation:

Given that  33.4% of people have sleepwalked.

Sample size n =1459

Sample favourable persons = 526

Sample proportion p = \frac{526}{1459} \\=0.361

Sample proportion p is normal for large samples with mean = 0.334 and

std error = \sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.334(1-0.334)}{1459} } \\=0.0123

a) P(526 or more of the 1459 adults have sleepwalked.)

=P(p\geq 0.361)\\=P(Z\geq \frac{0.361-0.334}{0.0123} \\=P(Z\geq 2.20)\\=0.5-0.4861\\=0.0139

b) Yes, because hardly 1.4% is the probability

c) 33.4 is very less compared to the average.  Either sample should be improved representing the population or population mean should be increased accordingly.

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Answer:

<em>The fraction of the flour Mrs. Mannng will use for her chocolate chip cookies is 4/9</em>

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<u>Fraction of Portion</u>

Suppose we have a number n and another number m such as m>n. The portion or fraction of size n of m is given by the quotient n/m.

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