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A college professor asked every student in his statistics class to flip a coin 100 times and count the number of heads. The results followed a normal distribution, with a mean of 50 and a standard deviation of 5.

If there were 70 students in the class, how many of the students most likely got heads between 45 and 55?

Answer to the nearest whole number of students.

[tex]\boxed{\text{students}}[/tex]

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Answer :

GinnyAnswer
Sure, let's solve this problem step-by-step.

### Step 1: Understand the Problem
Each student flips a coin 100 times. The number of heads follows a normal distribution:
- Mean ([tex]\(\mu\)[/tex]) = 50
- Standard Deviation ([tex]\(\sigma\)[/tex]) = 15

There are 70 students in total in the class.

We need to find how many students most likely got between 48 and 52 heads.

### Step 2: Calculate the Z-scores
The Z-score helps us to translate the number of heads (raw score) into a standard score under the normal distribution.

Formula for Z-score:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]

Where:
- [tex]\( X \)[/tex] is the raw score
- [tex]\( \mu \)[/tex] is the mean
- [tex]\( \sigma \)[/tex] is the standard deviation

Calculate Z-score for the lower bound (48 heads):
[tex]\[ Z_{lower} = \frac{48 - 50}{15} \][/tex]
[tex]\[ Z_{lower} = \frac{-2}{15} \][/tex]
[tex]\[ Z_{lower} = -0.1333 \][/tex]

Calculate Z-score for the upper bound (52 heads):
[tex]\[ Z_{upper} = \frac{52 - 50}{15} \][/tex]
[tex]\[ Z_{upper} = \frac{2}{15} \][/tex]
[tex]\[ Z_{upper} = 0.1333 \][/tex]

### Step 3: Find the Probability
We now need to find the probability that a student's Z-score falls between these two calculated Z-scores.

Using the standard normal distribution table or a cumulative distribution function (CDF):
1. Find the cumulative probability for [tex]\( Z_{upper} = 0.1333 \)[/tex]
2. Find the cumulative probability for [tex]\( Z_{lower} = -0.1333 \)[/tex]
3. Subtract the lower cumulative probability from the upper cumulative probability to get the probability of a student's number of heads falling between 48 and 52.

[tex]\[ P(48 \leq X \leq 52) = P(Z \leq 0.1333) - P(Z \leq -0.1333) \][/tex]
[tex]\[ P(48 \leq X \leq 52) = 0.5530 - 0.4467 \][/tex]
[tex]\[ P(48 \leq X \leq 52) = 0.1061 \][/tex]

### Step 4: Calculate the Number of Students
Now that we have the probability, we multiply it by the total number of students:

[tex]\[ \text{Number of students} = 0.1061 \times 70 \][/tex]
[tex]\[ \text{Number of students} \approx 7 \][/tex]

### Step 5: Conclusion
The number of students in the class who most likely got between 48 and 52 heads is approximately 7.

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