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Answer :
We are given that the time cats sleep each day follows a normal distribution with mean
[tex]$$\mu = 15 \text{ hours}$$[/tex]
and standard deviation
[tex]$$\sigma = 2 \text{ hours}.$$[/tex]
Let [tex]$X$[/tex] be the random variable representing the sleeping time. We want to find the probability that a cat sleeps between 13 hours and 17 hours, i.e., we want to calculate
[tex]$$P(13 \leq X \leq 17).$$[/tex]
Step 1. Standardize the values.
The standardization formula is
[tex]$$
z = \frac{x - \mu}{\sigma}.
$$[/tex]
For the lower bound ([tex]$x = 13$[/tex] hours):
[tex]$$
z_{\text{lower}} = \frac{13 - 15}{2} = \frac{-2}{2} = -1.
$$[/tex]
For the upper bound ([tex]$x = 17$[/tex] hours):
[tex]$$
z_{\text{upper}} = \frac{17 - 15}{2} = \frac{2}{2} = 1.
$$[/tex]
Step 2. Use the standard normal distribution.
The probability that [tex]$X$[/tex] is between 13 and 17 hours is the same as finding:
[tex]$$
P(13 \leq X \leq 17) = P(-1 \leq Z \leq 1).
$$[/tex]
This can be calculated using the cumulative distribution function (CDF) of the standard normal distribution, denoted by [tex]$\Phi(z)$[/tex]:
[tex]$$
P(-1 \leq Z \leq 1) = \Phi(1) - \Phi(-1).
$$[/tex]
Step 3. Find the probability.
Evaluating the values, we have:
[tex]$$
\Phi(1) \approx 0.8413,
$$[/tex]
and
[tex]$$
\Phi(-1) \approx 0.1587.
$$[/tex]
Then, the probability becomes:
[tex]$$
P(-1 \leq Z \leq 1) \approx 0.8413 - 0.1587 = 0.6827.
$$[/tex]
Thus, as a percentage:
[tex]$$
0.6827 \times 100\% \approx 68.27\%.
$$[/tex]
Conclusion
Approximately [tex]$68.27\%$[/tex] of the cats sleep between 13 hours and 17 hours.
[tex]$$\mu = 15 \text{ hours}$$[/tex]
and standard deviation
[tex]$$\sigma = 2 \text{ hours}.$$[/tex]
Let [tex]$X$[/tex] be the random variable representing the sleeping time. We want to find the probability that a cat sleeps between 13 hours and 17 hours, i.e., we want to calculate
[tex]$$P(13 \leq X \leq 17).$$[/tex]
Step 1. Standardize the values.
The standardization formula is
[tex]$$
z = \frac{x - \mu}{\sigma}.
$$[/tex]
For the lower bound ([tex]$x = 13$[/tex] hours):
[tex]$$
z_{\text{lower}} = \frac{13 - 15}{2} = \frac{-2}{2} = -1.
$$[/tex]
For the upper bound ([tex]$x = 17$[/tex] hours):
[tex]$$
z_{\text{upper}} = \frac{17 - 15}{2} = \frac{2}{2} = 1.
$$[/tex]
Step 2. Use the standard normal distribution.
The probability that [tex]$X$[/tex] is between 13 and 17 hours is the same as finding:
[tex]$$
P(13 \leq X \leq 17) = P(-1 \leq Z \leq 1).
$$[/tex]
This can be calculated using the cumulative distribution function (CDF) of the standard normal distribution, denoted by [tex]$\Phi(z)$[/tex]:
[tex]$$
P(-1 \leq Z \leq 1) = \Phi(1) - \Phi(-1).
$$[/tex]
Step 3. Find the probability.
Evaluating the values, we have:
[tex]$$
\Phi(1) \approx 0.8413,
$$[/tex]
and
[tex]$$
\Phi(-1) \approx 0.1587.
$$[/tex]
Then, the probability becomes:
[tex]$$
P(-1 \leq Z \leq 1) \approx 0.8413 - 0.1587 = 0.6827.
$$[/tex]
Thus, as a percentage:
[tex]$$
0.6827 \times 100\% \approx 68.27\%.
$$[/tex]
Conclusion
Approximately [tex]$68.27\%$[/tex] of the cats sleep between 13 hours and 17 hours.
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