Question

according to the transportation security administration (tsa) data, 6.3% of the passengers fly first - class. a random sample of size 176 was obtained. let $widehat{p}$ be the proportion of the sample that fly first - class. 1. use the central limit theorem (select an answer) to describe the probability distribution of $widehat{p}$ and state its parameters $mu_{widehat{p}}$ and $sigma_{widehat{p}}$: (round the answers to 4 decimal places) $widehat{p}sim$ (select an answer) ($mu_{widehat{p}}=$, $sigma_{widehat{p}}=$) 2. find the probability that between 4% and 9% of the sample fly first - class. (round the answer to 4 decimal places)

Explanation:

Step1: Identify population proportion and sample size

Let $p = 0.063$ (population proportion of first - class flyers) and $n=176$ (sample size).

Step2: Calculate mean of the sampling distribution of $\hat{p}$

By the Central Limit Theorem, $\mu_{\hat{p}}=p$. So $\mu_{\hat{p}} = 0.0630$.

Step3: Calculate standard deviation of the sampling distribution of $\hat{p}$

The formula for $\sigma_{\hat{p}}$ is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$. Substitute $p = 0.063$ and $n = 176$:

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The sampling distribution of $\hat{p}$ is approximately normal, i.e., $\hat{p}\sim N(\mu_{\hat{p}} = 0.0630,\sigma_{\hat{p}}\approx0.0183)$

Step4: Standardize the bounds for the probability calculation

For $p_1 = 0.04$:

$$z_1=\frac{0.04 - 0.0630}{0.0183}=\frac{- 0.023}{0.0183}\approx - 1.26$$

For $p_2 = 0.09$:

$$z_2=\frac{0.09 - 0.0630}{0.0183}=\frac{0.027}{0.0183}\approx1.48$$

Step5: Find the probability

$P(0.04<\hat{p}<0.09)=P(-1.26 < Z<1.48)=P(Z < 1.48)-P(Z < - 1.26)$
From the standard - normal table, $P(Z < 1.48)=0.9306$ and $P(Z < - 1.26)=0.1038$.
So $P(-1.26 < Z<1.48)=0.9306-0.1038 = 0.8268$

Answer:

1. $\hat{p}\sim N(\mu_{\hat{p}} = 0.0630,\sigma_{\hat{p}}\approx0.0183)$
2. $0.8268$