To solve this problem using the normal approximation to the binomial distribution, we need to first establish the parameters of the binomial distribution.
Given:
- \( n = 180 \) (number of households)
- \( p = 0.60 \) (probability of a household having a personal computer)
The mean \( \mu \) and standard deviation \( \sigma \) of the binomial distribution are given by:
\[ \mu = n \cdot p = 180 \cdot 0.60 = 108 \]
\[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{180 \cdot 0.60 \cdot 0.40} = \sqrt{180 \cdot 0.24} = \sqrt{43.2} \approx 6.561 \]
Next, we want to find the probability that the number of households with a personal computer, \( X \), is more than 60 but fewer than 100:
\[ P(60 < X < 100) \]
Using the normal approximation, we convert the binomial distribution to a normal distribution. We apply continuity correction since we are working with a discrete distribution.
So we look for:
\[ P(60 < X < 100) \approx P(60.5 < Y < 99.5) \]
where \( Y \) is a normal random variable with mean \( \mu = 108 \) and standard deviation \( \sigma \approx 6.561 \).
Now, we can standardize the values to find the corresponding z-scores:
- For \( Y = 60.5 \):
\[ z = \frac{60.5 - 108}{6.561} \approx \frac{-47.5}{6.561} \approx -7.24 \]
- For \( Y = 99.5 \):
\[ z = \frac{99.5 - 108}{6.561} \approx \frac{-8.5}{6.561} \approx -1.30 \]
Next, we use the standard normal distribution to find the probabilities corresponding to these z-scores.
Using standard normal distribution tables or a calculator:
\[ P(Z < -7.24) \approx 0 \quad \text{(very small probability)} \]
\[ P(Z < -1.30) \approx 0.0968 \]
Thus,
\[ P(60.5 < Y < 99.5) \approx P(Z < -1.30) - P(Z < -7.24) \approx 0.0968 - 0 \approx 0.0968 \]
Finally, rounding to the nearest thousandth, we find that the probability that more than 60 but fewer than 100 households have a personal computer is:
\[ \boxed{0.097} \]
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