Question

of the percent of students at a large high school are left - handed. a statistics teacher selects a random sample of 100 students and records ( x=) the number of left - handed students in the sample.
(1) calculate the mean and standard deviation of the approximate normal distribution.
mean = (square) students (do not round.)
standard deviation = (square) students (round to 2 decimal places.)
(2) use this normal distribution to calculate the probability that at least 15 of the students in the sample are left - handed.
probability = (square) (round to 4 decimal places. leave your answer in decimal form.)

Explanation:

Step1: Identify Distribution Type

This is a binomial distribution problem (count of left - handed students) with \(n = 100\) (sample size) and we assume the population proportion of left - handed students \(p\). Usually, the proportion of left - handed people in the population is around \(p=0.10\) (we'll use this as a common estimate if not given otherwise). For a binomial distribution \(X\sim B(n,p)\), the mean \(\mu=np\) and standard deviation \(\sigma=\sqrt{np(1 - p)}\).

Step2: Calculate the Mean

Given \(n = 100\) and assuming \(p = 0.10\) (since the problem is about left - handed students and a common proportion is 10%), the mean \(\mu=np=100\times0.10 = 10\) students.

Step3: Calculate the Standard Deviation

Using the formula for the standard deviation of a binomial distribution \(\sigma=\sqrt{np(1 - p)}\), with \(n = 100\), \(p = 0.10\) and \(1-p = 0.90\). So \(\sigma=\sqrt{100\times0.10\times0.90}=\sqrt{9}=3.00\) (rounded to 2 decimal places).

Step4: Use Normal Approximation

To use the normal approximation to the binomial, we check \(np\geq5\) and \(n(1 - p)\geq5\). Here \(np = 10\geq5\) and \(n(1 - p)=90\geq5\), so we can approximate \(X\sim B(100,0.10)\) with \(X\sim N(\mu = 10,\sigma = 3)\) (after continuity correction). We want \(P(X\geq15)\). First, apply continuity correction: \(P(X\geq15)=P(X>14.5)\) in the normal approximation.

Calculate the z - score: \(z=\frac{x-\mu}{\sigma}=\frac{14.5 - 10}{3}=\frac{4.5}{3}=1.5\)

Then, \(P(X>14.5)=1 - P(Z\leq1.5)\). Looking up \(P(Z\leq1.5)\) in the standard normal table, we get \(P(Z\leq1.5) = 0.9332\). So \(P(X\geq15)=1 - 0.9332=0.0668\) (rounded to 4 decimal places).

Answer:

• Mean: \(\boldsymbol{10}\) students
• Standard Deviation: \(\boldsymbol{3.00}\) students
• Probability: \(\boldsymbol{0.0668}\)