The probability that 75% or more of the women in the sample have been on a diet is 0.0371 or 3.71 %.
Here, n = 267, p = 0.7, 1 = 0.3
Mean, [tex]m_p = p[/tex]
⇒ [tex]m_p = 0.7[/tex]
Standard deviation, [tex]\sigma_p = \sqrt(\frac{pq}{n})[/tex]
⇒ [tex]\sigma_p[/tex] = 0.028
The conditions for a sampling distribution to be normal distribution, it must satisfy
1. Randomization condition(SRS)
2. logo condition
3. success/failure condition
Hence, the given sampling distribution in the problem statement is approximately normal distribution.
n×p = 267×0.7 = 186.9 ≥ 10
n×q = 267×0.3 = 80.1 ≥ 10
Norm cdf = (0.75, 999, 0.7, 0.028) ≈ 0.0371 ≈ 3.71%
To learn more about normal distribution
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