Question

step 1
to check whether the normal approximation to the binomial distribution is appropriate for this case, we confirm ( np > 5 ) and ( nq > 5 ), where sample size ( n = 85 ) and mean ( p = 0.8 ) are given, and ( q = 1 - p ).
find ( np ) and ( nq ).
( np = 85(0.8) )
( = 68 )
( nq = 85(1 - 0.8) )
( = 17 )
because ( np > 5 ) and ( nq > 5 ), the normal approximation to the binomial distribution is appropriate.
step 2
locate the mean, ( p = 0.8 ), and the range of observed values, ( 0.75 leq hat{p} leq 0.88 ), on the normal curve.
next, we convert each ( hat{p} ) to a ( z )-value using the formula below.
( z = \frac{hat{p} - p}{sqrt{\frac{pq}{n}}} )
recall ( n = 85 ), ( p = 0.8 ), and ( 0.75 leq hat{p} leq 0.88 ). calculate ( z_1 ) corresponding to ( hat{p} geq 0.75 ). round your answer to two decimal places.
( z_1 = \frac{hat{p}_1 - p}{sqrt{\frac{pq}{n}}} )
( = \frac{0.75 - 0.8}{sqrt{\frac{0.8(1 - 0.8)}{85}}} )
calculate ( z_2 ) corresponding to ( hat{p} leq 0.88 ). round your answer to two decimal places.
( z_2 = \frac{hat{p}_2 - p}{sqrt{\frac{pq}{n}}} )
( = \frac{0.88 - 0.8}{sqrt{\frac{0.8(1 - 0.8)}{85}}} )

Explanation:

Step1: Calculate the denominator for \( z_1 \)

First, we need to find the value of \( \sqrt{\frac{pq}{n}} \) where \( p = 0.8 \), \( q = 1 - p = 0.2 \), and \( n = 85 \).
\[
\frac{pq}{n}=\frac{0.8\times0.2}{85}=\frac{0.16}{85}\approx0.001882
\]
\[
\sqrt{\frac{pq}{n}}=\sqrt{0.001882}\approx0.0434
\]

Step2: Calculate the numerator for \( z_1 \)

The numerator is \( \hat{p}_1 - p \) where \( \hat{p}_1 = 0.75 \) and \( p = 0.8 \).
\[
\hat{p}_1 - p = 0.75 - 0.8=- 0.05
\]

Step3: Calculate \( z_1 \)

Now, we divide the numerator by the denominator to get \( z_1 \).
\[
z_1=\frac{-0.05}{0.0434}\approx - 1.15
\]

Step4: Calculate the denominator for \( z_2 \)

The denominator for \( z_2 \) is the same as for \( z_1 \) since \( p \), \( q \), and \( n \) are the same. So \( \sqrt{\frac{pq}{n}}\approx0.0434 \)

Step5: Calculate the numerator for \( z_2 \)

The numerator is \( \hat{p}_2 - p \) where \( \hat{p}_2 = 0.88 \) and \( p = 0.8 \).
\[
\hat{p}_2 - p = 0.88 - 0.8 = 0.08
\]

Step6: Calculate \( z_2 \)

Now, we divide the numerator by the denominator to get \( z_2 \).
\[
z_2=\frac{0.08}{0.0434}\approx1.84
\]

Answer:

s:
For \( z_1 \): \(-1.15\)
For \( z_2 \): \(1.84\)