Answer :
We are given that the daily temperature is normally distributed with mean
[tex]$$\mu = 69^\circ F$$[/tex]
and standard deviation
[tex]$$\sigma = 7^\circ F.$$[/tex]
To find the probability that the temperature is between [tex]$55^\circ F$[/tex] and [tex]$76^\circ F$[/tex], we first convert these temperatures to [tex]$z$[/tex]-scores using the formula
[tex]$$
z = \frac{x - \mu}{\sigma}.
$$[/tex]
1. For [tex]$x = 55^\circ F$[/tex]:
[tex]$$
z_{\text{lower}} = \frac{55 - 69}{7} = \frac{-14}{7} = -2.
$$[/tex]
2. For [tex]$x = 76^\circ F$[/tex]:
[tex]$$
z_{\text{upper}} = \frac{76 - 69}{7} = \frac{7}{7} = 1.
$$[/tex]
Next, we use the cumulative distribution function (CDF) of the standard normal distribution to find the probabilities corresponding to these [tex]$z$[/tex]-scores. Let [tex]$\Phi(z)$[/tex] denote the CDF.
- For [tex]$z = 1$[/tex], the probability is approximately
[tex]$$
\Phi(1) \approx 0.8413.
$$[/tex]
- For [tex]$z = -2$[/tex], the probability is approximately
[tex]$$
\Phi(-2) \approx 0.0228.
$$[/tex]
The probability that the temperature is between [tex]$55^\circ F$[/tex] and [tex]$76^\circ F$[/tex] is given by
[tex]$$
P(55 \leq T \leq 76) = \Phi(1) - \Phi(-2) \approx 0.8413 - 0.0228 = 0.8185.
$$[/tex]
To express this as a percentage, we multiply by 100:
[tex]$$
0.8185 \times 100 \approx 81.85\%.
$$[/tex]
Thus, about [tex]$81.86\%$[/tex] of the time, the temperature is between [tex]$55^\circ F$[/tex] and [tex]$76^\circ F$[/tex]. This corresponds to option D.
[tex]$$\mu = 69^\circ F$$[/tex]
and standard deviation
[tex]$$\sigma = 7^\circ F.$$[/tex]
To find the probability that the temperature is between [tex]$55^\circ F$[/tex] and [tex]$76^\circ F$[/tex], we first convert these temperatures to [tex]$z$[/tex]-scores using the formula
[tex]$$
z = \frac{x - \mu}{\sigma}.
$$[/tex]
1. For [tex]$x = 55^\circ F$[/tex]:
[tex]$$
z_{\text{lower}} = \frac{55 - 69}{7} = \frac{-14}{7} = -2.
$$[/tex]
2. For [tex]$x = 76^\circ F$[/tex]:
[tex]$$
z_{\text{upper}} = \frac{76 - 69}{7} = \frac{7}{7} = 1.
$$[/tex]
Next, we use the cumulative distribution function (CDF) of the standard normal distribution to find the probabilities corresponding to these [tex]$z$[/tex]-scores. Let [tex]$\Phi(z)$[/tex] denote the CDF.
- For [tex]$z = 1$[/tex], the probability is approximately
[tex]$$
\Phi(1) \approx 0.8413.
$$[/tex]
- For [tex]$z = -2$[/tex], the probability is approximately
[tex]$$
\Phi(-2) \approx 0.0228.
$$[/tex]
The probability that the temperature is between [tex]$55^\circ F$[/tex] and [tex]$76^\circ F$[/tex] is given by
[tex]$$
P(55 \leq T \leq 76) = \Phi(1) - \Phi(-2) \approx 0.8413 - 0.0228 = 0.8185.
$$[/tex]
To express this as a percentage, we multiply by 100:
[tex]$$
0.8185 \times 100 \approx 81.85\%.
$$[/tex]
Thus, about [tex]$81.86\%$[/tex] of the time, the temperature is between [tex]$55^\circ F$[/tex] and [tex]$76^\circ F$[/tex]. This corresponds to option D.
