Respuesta :
Answer:
The probability that no more than 50% in the sample of 700 had received such an email is 0.9830.
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n ≥ 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
[tex]\mu_{\hat p}=p[/tex]
The standard deviation of this sampling distribution of sample proportion is:
[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}[/tex]
The information provided is:
p = 54% = 0.54
n = 700
Since the sample size selected is quite large, i.e. n = 700 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion.
[tex]\hat p\sim N(p, \frac{p(1-p)}{n})[/tex]
Compute the probability that no more than 50% in the sample of 700 had received such an email as follows:
[tex]P(\hat p>0.50)=P(\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}>\frac{0.50-0.54}{\sqrt{\frac{0.54(1-0.54)}{700}}})[/tex]
[tex]=P(Z>-2.12)\\=P(Z<2.12)\\=0.9830[/tex]
*use a z-table for the probability.
Thus, the probability that no more than 50% in the sample of 700 had received such an email is 0.9830.
