Question

suppose percent of students at a large high school are left - handed. a statistician takes a random sample of 100 students and records ( x =) the number of left - handed students in the sample.
(a) calculate the mean and standard deviation of the appropriate normal distribution.
mean = (square) students (do not round.)
standard deviation = (square) students (round to 2 decimal places.)
(b) 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 (left - handed students: success/failure, fixed trials \(n = 100\), probability \(p\) of left - handed, assume \(p=0.1\) (common left - handed rate), but wait, the problem says "the percent of students at a large high school are left - handed" – maybe we missed the percentage? Wait, maybe it's a typo, but usually, the proportion of left - handed people is about \(0.1\) (10%). Let's assume \(p = 0.1\), \(n=100\). For normal approximation to binomial, \(np\geq5\) and \(n(1 - p)\geq5\). Here, \(np = 100\times0.1=10\), \(n(1 - p)=100\times0.9 = 90\), so normal approximation is valid.

Step2: Calculate Mean (\(\mu\))

For binomial distribution, mean \(\mu=np\). If \(n = 100\) and let's say \(p = 0.1\) (since it's a common left - handed proportion, maybe the problem had a typo and the percentage is 10%), then \(\mu=100\times0.1 = 10\) students.

Step3: Calculate Standard Deviation (\(\sigma\))

Standard deviation for binomial is \(\sigma=\sqrt{np(1 - p)}\). With \(n = 100\), \(p = 0.1\), \(\sigma=\sqrt{100\times0.1\times0.9}=\sqrt{9}=3\) students (rounded to 2 decimal places, it's 3.00).

Step4: Normal Approximation for Probability

We want \(P(X\geq15)\). For normal approximation, we use continuity correction, so \(P(X\geq15)=P(X>14.5)\) (since for discrete to continuous, we add/subtract 0.5). First, calculate the z - score: \(z=\frac{x-\mu}{\sigma}=\frac{14.5 - 10}{3}=\frac{4.5}{3}=1.5\). Then, \(P(Z>1.5)=1 - P(Z\leq1.5)\). From z - table, \(P(Z\leq1.5)=0.9332\), so \(P(Z>1.5)=1 - 0.9332 = 0.0668\). Wait, but if the proportion \(p\) is different, say if the original problem had a different \(p\), but since the problem is about normal approximation to binomial (which is a part of statistics), let's proceed with the steps.

Wait, maybe the problem had a percentage, like 15%? Wait, the image is a bit unclear. But assuming the standard left - handed proportion \(p = 0.1\) (10%):

Step1 (Revised if \(p\) is different):

Wait, maybe the problem's "percent of students" was, say, 15%? Let's check. If \(p = 0.15\), then \(\mu=np=100\times0.15 = 15\), \(\sigma=\sqrt{100\times0.15\times0.85}=\sqrt{12.75}\approx3.57\). But the problem says "calculate the mean and standard deviation of the appropriate normal distribution" for \(n = 100\) and \(T\) (number of left - handed). Let's assume the proportion \(p\) is given (maybe a typo in the image, but let's use the standard approach).

Assuming \(p = 0.1\) (10% left - handed):

• Mean (\(\mu\)): \(np=100\times0.1 = 10\)
• Standard Deviation (\(\sigma\)): \(\sqrt{np(1 - p)}=\sqrt{100\times0.1\times0.9}=3\)

For part (b), \(P(X\geq15)\) with continuity correction \(X>14.5\):

\(z=\frac{14.5 - 10}{3}=1.5\)

\(P(Z>1.5)=1 - 0.9332 = 0.0668\)

Answer:

(a) Mean = 10 students, Standard Deviation = 3.00 students (assuming \(p = 0.1\))
(b) Probability = 0.0668 (assuming \(p = 0.1\))

(Note: If the proportion \(p\) is different, the values will change. For example, if \(p = 0.15\), mean = 15, standard deviation=\(\sqrt{100\times0.15\times0.85}\approx3.57\), and \(z=\frac{14.5 - 15}{3.57}\approx - 0.14\), \(P(Z\geq - 0.14)=1 - P(Z < - 0.14)=1 - 0.4443 = 0.5557\). But since the problem's percentage of left - handed students is not clearly visible in the image, we used the common \(p = 0.1\) for illustration.)