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A distribution of values is normal with a mean of 213.8 and a standard deviation of 75.6. Find the probability that a randomly selected value is between 349.9 and 410.4.

[tex]\[ P(349.9 \ \textless \ X \ \textless \ 410.4) = \square \][/tex]

Sagot :

To find the probability that a randomly selected value from a normal distribution is between 349.9 and 410.4, given the mean (213.8) and standard deviation (75.6), we will follow these steps:

### Step 1: Calculate the Z-scores
First, we need to standardize the values 349.9 and 410.4 by converting them to Z-scores. The Z-score for a value [tex]\( X \)[/tex] in a normal distribution can be calculated using the formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\(\mu\)[/tex] is the mean and [tex]\(\sigma\)[/tex] is the standard deviation.

#### Z-score for the lower bound (349.9)
[tex]\[ Z_{\text{lower}} = \frac{349.9 - 213.8}{75.6} \][/tex]
[tex]\[ Z_{\text{lower}} \approx 1.800 \][/tex]

#### Z-score for the upper bound (410.4)
[tex]\[ Z_{\text{upper}} = \frac{410.4 - 213.8}{75.6} \][/tex]
[tex]\[ Z_{\text{upper}} \approx 2.601 \][/tex]

### Step 2: Use the Cumulative Distribution Function (CDF)
Next, we use the cumulative distribution function of the standard normal distribution to find the probabilities corresponding to these Z-scores.

Let [tex]\( \Phi(Z) \)[/tex] denote the CDF of the standard normal distribution.

#### Probability corresponding to [tex]\( Z_{\text{lower}} \)[/tex]
[tex]\[ \Phi(1.800) \approx 0.964 \][/tex]

#### Probability corresponding to [tex]\( Z_{\text{upper}} \)[/tex]
[tex]\[ \Phi(2.601) \approx 0.995 \][/tex]

### Step 3: Find the Probability Between the Two Z-scores
To find the probability that a value lies between our two Z-scores, we subtract the smaller CDF value from the larger CDF value:
[tex]\[ P(1.800 < Z < 2.601) = \Phi(2.601) - \Phi(1.800) \][/tex]
[tex]\[ P(1.800 < Z < 2.601) \approx 0.995 - 0.964 \][/tex]
[tex]\[ P(1.800 < Z < 2.601) \approx 0.031 \][/tex]

Thus, the probability that a randomly selected value from the normal distribution with a mean of 213.8 and a standard deviation of 75.6 is between 349.9 and 410.4 is approximately [tex]\( 0.031 \)[/tex] or [tex]\( 3.125\% \)[/tex].

Therefore,
[tex]\[ P(349.9 < X < 410.4) \approx 0.031 \][/tex]