Answer :
The probability that no more than [tex]25[/tex] were victims of e-mail fraud is [tex]\fbox{0.00169}[/tex].
Further explanation:
Given:
The probability of a user experience e-mail fraud [tex]p[/tex] is [tex]0.7[/tex].
The number of individuals [tex]n[/tex] are [tex]50[/tex].
Calculation:
The [tex]\bar{X}[/tex] follow the Binomial distribution can be expressed as,
[tex]\bar{X}\sim \text{Binomial}(n,p)[/tex]
Use the normal approximation for [tex]\bar{X}[/tex] as
[tex]\bar{X}\sim \text{Normal}(np,np(1-p))[/tex]
The mean [tex]\mu[/tex] is [tex]\fbox{np}[/tex]
The standard deviation [tex]\sigma[/tex] is [tex]\fbox{\begin{minispace}\\ \sqrt{np(1-p)}\end{minispace}}[/tex]
The value of [tex]\mu[/tex] can be calculated as,
[tex]\mu=np\\ \mu= 50 \times0.7\\ \mu=35[/tex]
The value of [tex]\sigma[/tex] can be calculated as,
[tex]\sigma=\sqrt{50\times0.7\times(1-0.7)} \\\sigma=\sqrt{50\times0.7\times0.3}\\\sigma=\sqrt{10.5}[/tex]
By Normal approximation \bar{X} also follow Normal distribution as,
[tex]\bar{X}\sim \text{Normal}(\mu,\sigma^{2} )[/tex]
Substitute 35 for [tex]\mu[/tex] and 10.5 for [tex]\sigma^{2}[/tex]
[tex]\bar{X}\sim\text {Normal}(35,10.5)[/tex]
The probability that not more than [tex]25[/tex] were victims of e-mail fraud can be calculated as,
[tex]\text{Probability}=P(\bar{X}<25)}\\\text{Probability}=P(\frac{{\bar{X}-\mu}}{\sigma}<\frac{{(25+0.5)-35}}{\sqrt{10.5} })\\\text{Probability}=P(Z}<\frac{{25.5-35}}{\sqrt{10.5} })\\\text{Probability}=P(Z}<-2.93})\\[/tex]
The Normal distribution is symmetric.
[tex]P(Z>-2.93})=1-P(Z<2.93)\\P(Z>-2.93})=1-0.99831\\P(Z>-2.93})=0.00169[/tex]
Hence, the probability that no more than [tex]25[/tex] were victims of e-mail fraud is [tex]\fbox{0.00169}[/tex].
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Answer Details:
Grade: College Statistics
Subject: Mathematics
Chapter: Probability and Statistics
Keywords:
Probability, Statistics, E-mail fraud, internet, Binomial distribution, Normal distribution, Normal approximation, Central Limit Theorem, Z-table, Mean, Standard deviation, Symmetric.