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Answer :
We are given that the annual rainfall, [tex]$X$[/tex], is normally distributed with a mean [tex]$$\mu = 68$$[/tex] inches and a standard deviation [tex]$$\sigma = 3$$[/tex] inches. We wish to find the probability that the rainfall is between 68 inches and 76 inches, i.e., we want to calculate
[tex]$$
P(68 \le X \le 76).
$$[/tex]
Step 1. Standardization
To find this probability we first convert the values to [tex]$z$[/tex]-scores using the standardization formula
[tex]$$
z = \frac{x - \mu}{\sigma}.
$$[/tex]
For the lower bound, when [tex]$x = 68$[/tex], the [tex]$z$[/tex]-score is
[tex]$$
z_{\text{lower}} = \frac{68 - 68}{3} = 0.
$$[/tex]
For the upper bound, when [tex]$x = 76$[/tex], the [tex]$z$[/tex]-score is
[tex]$$
z_{\text{upper}} = \frac{76 - 68}{3} = \frac{8}{3} \approx 2.6667.
$$[/tex]
Step 2. Finding the Corresponding Probabilities
Under the standard normal distribution, let [tex]$\Phi(z)$[/tex] be the cumulative distribution function (CDF) which gives the probability that a standard normal variable is less than or equal to [tex]$z$[/tex].
- For [tex]$z = 0$[/tex], we have
[tex]$$
\Phi(0) = 0.5.
$$[/tex]
- For [tex]$z \approx 2.6667$[/tex], the cumulative probability is approximately
[tex]$$
\Phi(2.6667) \approx 0.99617.
$$[/tex]
Step 3. Calculating the Desired Probability
The probability that [tex]$X$[/tex] is between 68 and 76 inches corresponds to the probability that [tex]$Z$[/tex] is between 0 and 2.6667. This can be calculated as
[tex]$$
P(0 \le Z \le 2.6667) = \Phi(2.6667) - \Phi(0).
$$[/tex]
Substituting the values we found,
[tex]$$
P(68 \le X \le 76) = 0.99617 - 0.5 \approx 0.49617.
$$[/tex]
Rounding to four decimal places, we obtain
[tex]$$
P(68 \le X \le 76) \approx 0.4962.
$$[/tex]
Final Answer
The probability that the annual rainfall is between 68 inches and 76 inches is approximately [tex]$$0.4962.$$[/tex]
[tex]$$
P(68 \le X \le 76).
$$[/tex]
Step 1. Standardization
To find this probability we first convert the values to [tex]$z$[/tex]-scores using the standardization formula
[tex]$$
z = \frac{x - \mu}{\sigma}.
$$[/tex]
For the lower bound, when [tex]$x = 68$[/tex], the [tex]$z$[/tex]-score is
[tex]$$
z_{\text{lower}} = \frac{68 - 68}{3} = 0.
$$[/tex]
For the upper bound, when [tex]$x = 76$[/tex], the [tex]$z$[/tex]-score is
[tex]$$
z_{\text{upper}} = \frac{76 - 68}{3} = \frac{8}{3} \approx 2.6667.
$$[/tex]
Step 2. Finding the Corresponding Probabilities
Under the standard normal distribution, let [tex]$\Phi(z)$[/tex] be the cumulative distribution function (CDF) which gives the probability that a standard normal variable is less than or equal to [tex]$z$[/tex].
- For [tex]$z = 0$[/tex], we have
[tex]$$
\Phi(0) = 0.5.
$$[/tex]
- For [tex]$z \approx 2.6667$[/tex], the cumulative probability is approximately
[tex]$$
\Phi(2.6667) \approx 0.99617.
$$[/tex]
Step 3. Calculating the Desired Probability
The probability that [tex]$X$[/tex] is between 68 and 76 inches corresponds to the probability that [tex]$Z$[/tex] is between 0 and 2.6667. This can be calculated as
[tex]$$
P(0 \le Z \le 2.6667) = \Phi(2.6667) - \Phi(0).
$$[/tex]
Substituting the values we found,
[tex]$$
P(68 \le X \le 76) = 0.99617 - 0.5 \approx 0.49617.
$$[/tex]
Rounding to four decimal places, we obtain
[tex]$$
P(68 \le X \le 76) \approx 0.4962.
$$[/tex]
Final Answer
The probability that the annual rainfall is between 68 inches and 76 inches is approximately [tex]$$0.4962.$$[/tex]
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