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according to a survey, 10% of americans are afraid to fly. suppose 1,100 americans are sampled.
a. what is the probability percentage that 121 or more americans in the survey are afraid to fly? round
the percent to two decimal places. 14.47 ×
enter an integer or decimal number (more..)
b. what is the probability percentage that 165 or more americans in the survey are afraid to fly? round
the percent to two decimal places. 0.00 ✔
c. what is the probability percentage that 8% or less of the americans surveyed answered they were
afraid to fly? round the percent to two decimal places. 1.50 ×

Explanation:

Part (a)

This is a binomial probability problem, but we can use the normal approximation to the binomial distribution since \( n \) is large (\( n = 1100 \)), \( p = 0.1 \), so \( np = 1100\times0.1 = 110 \), \( n(1 - p)=1100\times0.9 = 990 \), both are greater than 5.

The mean of the binomial distribution (and the normal approximation) is \( \mu = np = 110 \), and the standard deviation is \( \sigma=\sqrt{np(1 - p)}=\sqrt{1100\times0.1\times0.9}=\sqrt{99}\approx9.95 \)

We want \( P(X\geq121) \). For normal approximation, we use continuity correction, so we find \( P(X\geq120.5) \)

First, calculate the z - score: \( z=\frac{x-\mu}{\sigma}=\frac{120.5 - 110}{9.95}=\frac{10.5}{9.95}\approx1.055 \)

We want \( P(Z\geq1.055) \). Since \( P(Z\geq z)=1 - P(Z\lt z) \), and from the standard normal table, \( P(Z\lt1.06)\approx0.8554 \) (using more precise calculation: \( P(Z\lt1.055)\approx0.8547 \))

So \( P(Z\geq1.055)=1 - 0.8547 = 0.1453 \), which is \( 14.53\% \) (rounded to two decimal places)

Step 1: Identify the distribution and parameters

We have a binomial distribution with \( n = 1100 \), \( p = 0.1 \). We use normal approximation because \( np\) and \( n(1 - p)\) are large.
Mean \( \mu=np = 1100\times0.1 = 110 \)
Standard deviation \( \sigma=\sqrt{np(1 - p)}=\sqrt{1100\times0.1\times0.9}=\sqrt{99}\approx9.95 \)

Step 2: Apply continuity correction

For \( P(X\geq121) \), with continuity correction, we consider \( X\geq120.5 \)

Step 3: Calculate the z - score

\( z=\frac{120.5 - 110}{9.95}=\frac{10.5}{9.95}\approx1.06 \) (rounded to two decimal places)

Step 4: Find the probability from the z - score

\( P(Z\geq1.06)=1 - P(Z\lt1.06) \)
From standard normal table, \( P(Z\lt1.06) = 0.8554 \)
So \( P(Z\geq1.06)=1 - 0.8554 = 0.1446\approx14.46\% \) (more accurately, if we use \( z = 1.055 \), \( P(Z\lt1.055)\approx0.8547 \), \( P(Z\geq1.055)=1 - 0.8547 = 0.1453\approx14.53\% \))

Part (c)

First, find the number of Americans corresponding to 8% of 1100: \( x = 0.08\times1100 = 88 \)

We want \( P(X\leq88) \). Using continuity correction, we find \( P(X\leq88.5) \)

The mean \( \mu = 110 \), standard deviation \( \sigma\approx9.95 \)

Calculate the z - score: \( z=\frac{88.5 - 110}{9.95}=\frac{- 21.5}{9.95}\approx - 2.16 \)

We want \( P(Z\leq - 2.16) \). From the standard normal table, \( P(Z\leq - 2.16)=1 - P(Z\lt2.16) \)

\( P(Z\lt2.16) = 0.9846 \), so \( P(Z\leq - 2.16)=1 - 0.9846 = 0.0154\approx1.54\% \) (rounded to two decimal places)

Step 1: Find the number of successes for 8%

\( x = 0.08\times1100 = 88 \)

Step 2: Apply continuity correction

For \( P(X\leq88) \), with continuity correction, we consider \( X\leq88.5 \)

Step 3: Calculate the z - score

\( z=\frac{88.5 - 110}{9.95}=\frac{-21.5}{9.95}\approx - 2.16 \)

Step 4: Find the probability from the z - score

\( P(Z\leq - 2.16)=1 - P(Z\lt2.16) \)
From standard normal table, \( P(Z\lt2.16)=0.9846 \)
So \( P(Z\leq - 2.16)=1 - 0.9846 = 0.0154\approx1.54\% \)

Answer:

s:
a. \( \approx14.53\% \) (or 14.46% depending on z - score precision)
c. \( \approx1.54\% \)