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To solve this problem, we need to determine the probability of flipping 48 or 49 heads out of 88 flips. We are told that the number of heads, \(X\), follows an approximate normal distribution \(N(44, 4.7)\), where the mean \(\mu\) is 44 and the standard deviation \(\sigma\) is 4.7. Here's how we can solve it step by step:
1. Convert the given number of heads (48 and 49) into z-scores.
The z-score formula is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
- For \(X = 48\):
[tex]\[ z_{48} = \frac{48 - 44}{4.7} = \frac{4}{4.7} \approx 0.8511 \][/tex]
- For \(X = 49\):
[tex]\[ z_{49} = \frac{49 - 44}{4.7} = \frac{5}{4.7} \approx 1.0638 \][/tex]
2. Find the cumulative probability associated with each z-score using the standard normal table.
- For \(z_{48} = 0.8511\):
- To find the probability, we look up the closest value for \(z = 0.85\) in the table.
[tex]\[ P(Z \leq 0.85) \approx 0.8023 \][/tex]
- Since 0.8511 is closer to 0.85 than 0.86, we approximate:
[tex]\[ P(Z \leq 0.8511) \approx 0.8026 \][/tex]
- For \(z_{49} = 1.0638\):
- To find the probability, we look up the closest value for \(z = 1.06\) in the table.
[tex]\[ P(Z \leq 1.06) \approx 0.8554 \][/tex]
- Since 1.0638 is closer to 1.06 than 1.07, we approximate:
[tex]\[ P(Z \leq 1.0638) \approx 0.8563 \][/tex]
3. Calculate the probability of flipping between 48 and 49 heads.
The probability of flipping between 48 and 49 heads is the difference in the cumulative probabilities:
[tex]\[ P(48 \leq X \leq 49) = P(Z \leq 1.0638) - P(Z \leq 0.8511) \][/tex]
Substitute the values:
[tex]\[ P(48 \leq X \leq 49) \approx 0.8563 - 0.8026 = 0.0537 \][/tex]
4. Round the final answer to two decimal places:
[tex]\[ \text{Rounded probability} \approx 0.05 \][/tex]
Therefore, the probability of flipping 48 or 49 heads out of 88 flips is approximately [tex]\(0.05\)[/tex].
1. Convert the given number of heads (48 and 49) into z-scores.
The z-score formula is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
- For \(X = 48\):
[tex]\[ z_{48} = \frac{48 - 44}{4.7} = \frac{4}{4.7} \approx 0.8511 \][/tex]
- For \(X = 49\):
[tex]\[ z_{49} = \frac{49 - 44}{4.7} = \frac{5}{4.7} \approx 1.0638 \][/tex]
2. Find the cumulative probability associated with each z-score using the standard normal table.
- For \(z_{48} = 0.8511\):
- To find the probability, we look up the closest value for \(z = 0.85\) in the table.
[tex]\[ P(Z \leq 0.85) \approx 0.8023 \][/tex]
- Since 0.8511 is closer to 0.85 than 0.86, we approximate:
[tex]\[ P(Z \leq 0.8511) \approx 0.8026 \][/tex]
- For \(z_{49} = 1.0638\):
- To find the probability, we look up the closest value for \(z = 1.06\) in the table.
[tex]\[ P(Z \leq 1.06) \approx 0.8554 \][/tex]
- Since 1.0638 is closer to 1.06 than 1.07, we approximate:
[tex]\[ P(Z \leq 1.0638) \approx 0.8563 \][/tex]
3. Calculate the probability of flipping between 48 and 49 heads.
The probability of flipping between 48 and 49 heads is the difference in the cumulative probabilities:
[tex]\[ P(48 \leq X \leq 49) = P(Z \leq 1.0638) - P(Z \leq 0.8511) \][/tex]
Substitute the values:
[tex]\[ P(48 \leq X \leq 49) \approx 0.8563 - 0.8026 = 0.0537 \][/tex]
4. Round the final answer to two decimal places:
[tex]\[ \text{Rounded probability} \approx 0.05 \][/tex]
Therefore, the probability of flipping 48 or 49 heads out of 88 flips is approximately [tex]\(0.05\)[/tex].
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