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.
