Question

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}}}
= -1.15
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}}}
= 1.84
step 3
now that we have ( z_1 = -1.15 ) and ( z_2 = 1.84 ), or ( p(-1.15 leq z leq 1.84) ), corresponding to ( p(0.75 leq hat{p} leq 0.88) ), we can calculate the probability.
recall that ( p(-1.15 leq z leq 1.84) ) can be rewritten ( p(z leq 1.84) - p(z leq -1.15) ). use salt to find the probability for ( p(z leq 1.84) ). round your answer to four decimal places.
( p(z leq 1.84) = )
use salt to find the probability for ( p(z leq -1.15) ). round your answer to four decimal places.
( p(z leq -1.15) = )

Explanation:

To find \( P(z \leq 1.84) \) and \( P(z \leq -1.15) \), we use the standard normal distribution table (or a calculator with normal distribution functionality).

Step 1: Find \( P(z \leq 1.84) \)

Using the standard normal distribution, we look up the z - score of 1.84. The cumulative probability for \( z = 1.84 \) is the area to the left of \( z = 1.84 \) under the standard normal curve. From the standard normal table or using a calculator, we find that \( P(z \leq 1.84)=0.9671\)

Step 2: Find \( P(z \leq - 1.15) \)

For a z - score of - 1.15, we find the cumulative probability (area to the left of \( z=-1.15 \)) from the standard normal table or calculator. The value of \( P(z \leq - 1.15) = 0.1251\)

Answer:

s:
\( P(z \leq 1.84)=\boxed{0.9671} \)

\( P(z \leq - 1.15)=\boxed{0.1251} \)