Using this approximation, find the probability of flipping 21 or 22 heads. You may use the portion of the Standard Normal Table below.

\begin{tabular}{c|cccccccccc}
[tex]$z$[/tex] & [tex]$0 . 0 0$[/tex] & [tex]$0 . 0 1$[/tex] & [tex]$0 . 0 2$[/tex] & [tex]$0 . 0 3$[/tex] & [tex]$0 . 0 4$[/tex] & [tex]$0 . 0 5$[/tex] & [tex]$0 . 0 6$[/tex] & [tex]$0 . 0 7$[/tex] & [tex]$0 . 0 8$[/tex] & [tex]$0 . 0 9$[/tex] \\
\hline [tex]$0 . 7$[/tex] & 0.7580 & 0.7611 & 0.7642 & 0.7673 & 0.7704 & 0.7734 & 0.7764 & 0.7794 & 0.7823 & 0.7852 \\
[tex]$0 . 8$[/tex] & 0.7881 & 0.7910 & 0.7939 & 0.7967 & 0.7995 & 0.8023 & 0.8051 & 0.8078 & 0.8106 & 0.8133 \\
[tex]$0 . 9$[/tex] & 0.8159 & 0.8186 & 0.8212 & 0.8238 & 0.8264 & 0.8289 & 0.8315 & 0.8340 & 0.8365 & 0.8389 \\
[tex]$1 . 0$[/tex] & 0.8413 & 0.8438 & 0.8461 & 0.8485 & 0.8508 & 0.8531 & 0.8554 & 0.8577 & 0.8599 & 0.8621 \\
[tex]$1 . 1$[/tex] & 0.8643 & 0.8665 & 0.8686 & 0.8708 & 0.8729 & 0.8749 & 0.8770 & 0.8790 & 0.8810 & 0.8830 \\
[tex]$1 . 2$[/tex] & 0.8849 & 0.8869 & 0.8888 & 0.8907 & 0.8925 & 0.8944 & 0.8962 & 0.8980 & 0.8997 & 0.9015 \\
[tex]$1 . 3$[/tex] & 0.9032 & 0.9049 & 0.9066 & 0.9082 & 0.9099 & 0.9115 & 0.9131 & 0.9147 & 0.9162 & 0.9177 \\
[tex]$1 . 4$[/tex] & 0.9192 & 0.9207 & 0.9222 & 0.9236 & 0.9251 & 0.9265 & 0.9279 & 0.9292 & 0.9306 & 0.9319 \\
[tex]$1 . 5$[/tex] & 0.9332 & 0.9345 & 0.9357 & 0.9370 & 0.9382 & 0.9394 & 0.9406 & 0.9418 & 0.9429 & 0.9441
\end{tabular}

Round the final answer to two decimal places.



Answer :

Certainly! Let's walk through the necessary steps to find the probability of flipping 21 or 22 heads, given the standard normal distribution and the provided Standard Normal Table.

### Step 1: Setup and Parameters
We need to use the normal approximation to the binomial distribution to find the probability of getting 21 or 22 heads out of 40 flips:
- The probability of heads (p) = 0.5
- The total number of flips (n) = 40

### Step 2: Mean and Standard Deviation
The mean (\(\mu\)) and standard deviation (\(\sigma\)) of a binomial distribution can be calculated using:
[tex]\[ \mu = n \times p = 40 \times 0.5 = 20 \][/tex]
[tex]\[ \sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{40 \times 0.5 \times 0.5} = \sqrt{10} \approx 3.16 \][/tex]

### Step 3: Continuity Correction and Z-scores
To find the probability of exactly 21 or 22 heads, we use the continuity correction.

For 21 heads, we use 21.5:
[tex]\[ z_{21.5} = \frac{21.5 - \mu}{\sigma} = \frac{21.5 - 20}{3.16} \approx 0.47 \][/tex]

For 22 heads, we use 22.5:
[tex]\[ z_{22.5} = \frac{22.5 - \mu}{\sigma} = \frac{22.5 - 20}{3.16} \approx 0.79 \][/tex]

### Step 4: Using the Standard Normal Table
From the provided Standard Normal Table, we need to find the cumulative probabilities for these z-scores:

For \(z \approx 0.47\):
[tex]\[ P(Z \leq 0.47) \approx 0.6794 \][/tex]

For \(z \approx 0.79\):
[tex]\[ P(Z \leq 0.79) \approx 0.7852 \][/tex]

### Step 5: Calculate the Desired Probability
The probability of flipping 21 or 22 heads is the difference between these cumulative probabilities.
[tex]\[ P(21 \leq X \leq 22) = P(Z \leq 0.79) - P(Z \leq 0.47) = 0.7852 - 0.6794 = 0.1058 \][/tex]

### Final Answer
Rounding the result to two decimal places:
[tex]\[ P(21 \leq X \leq 22) \approx 0.11 \][/tex]
Thus, the probability of flipping 21 or 22 heads is approximately 0.11.