Respuesta :
Answer:
Approximately [tex]0.60[/tex] (which is equivalent to [tex]60\%[/tex].)
Step-by-step explanation:
Here's how to find this probability using a [tex]Z[/tex]-table.
Let [tex]X[/tex] denote the price (in dollars) of one laptop price from this set of prices.
The question states that this set of prices is normally distributed with a mean of [tex]\$750[/tex] and a standard deviation of [tex]\$60[/tex]. Therefore, [tex]X[/tex] would be a normal random variable with mean [tex]\mu = 750[/tex] and standard deviation [tex]\sigma = 60[/tex]. .
What this question is asking for is (the same as) the probability that [tex]X[/tex] is between the lower bound [tex]624[/tex] and the upper bound [tex]768[/tex]. That is:
[tex]P(624 \le X \le 768)[/tex].
This probability can be written as the difference between of two simpler probabilities: the probability that [tex]X[/tex] is less than the upper bound [tex]P(X < 768)[/tex], minus the probability that [tex]X\![/tex] is less than the lower bound [tex]P(X < 624)[/tex].
The first probability [tex]P(X < 768)[/tex] covers a wide range of possibilities; if [tex]X < 624[/tex], then it must be true that [tex]X < 768[/tex]. Therefore, the region corresponding to the second probability [tex]P(X < 624)[/tex] is completely enclosed in the first region. Subtracting the second probability from the first will give the probability of the area between the upper and lower bound, which is indeed equal to [tex]P(624 \le X \le 768)[/tex].
[tex]P(624 \le X \le 768) = P(X \le 768) - P(X \le 624)[/tex].
Both [tex]P(X < 768)[/tex] and [tex]P(X < 624)[/tex] can be found using a [tex]Z[/tex]-table. Here are the steps:
Calculate the [tex]z[/tex]-score of the two bounds ([tex]624[/tex] and [tex]768[/tex]) with respect to the distribution of [tex]X[/tex], the laptop price. For a normal distribution of mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a measurement of value [tex]x[/tex] will have a [tex]z[/tex]-score of:
[tex]\displaystyle z = \frac{x - \mu}{\sigma}[/tex].
For the two bounds in this question:
- The lower bound, [tex]x = 624[/tex], will have a [tex]z[/tex]-score of [tex]z = \displaystyle \frac{624 - 750}{60} = -2.1[/tex].
- The upper bound, [tex]x = 768[/tex], will have a [tex]z[/tex]-score of [tex]z = \displaystyle \frac{768 - 750}{60} = 0.3[/tex].
Rewrite the two probabilities using the [tex]z[/tex]-score of the two bounds:
- [tex]P(X < 624)[/tex] would become [tex]P(Z < -2.1)[/tex].
- [tex]P(X < 768)[/tex] would become [tex]P(Z < 0.3)[/tex].
Look up the probabilities corresponding to these two [tex]z[/tex]-scores on a [tex]z\![/tex]-score table. Note, that some [tex]z\!\![/tex]-tables did not include the section where [tex]z < 0[/tex]. That's not a problem because the [tex]z\!\!\![/tex]-distribution is symmetric.
[tex]\begin{aligned}P(Z < -2.1) &= P(Z > 2.1) && \text{By the symmetry of $z$-distribution} \\ &= 1 - P(Z< 2.1)\end{aligned}[/tex].
On a typical [tex]z[/tex]-table:
- [tex]P(Z < 2.10) \approx 0.982136[/tex].
- [tex]P(Z < 0.30) \approx 0.617911[/tex].
Therefore:
[tex]\begin{aligned}P(X < 624) &= P(Z < -2.1)\\ &= 1 - P(Z < 2.1) \approx (1 - 0.982136) \end{aligned}[/tex].
[tex]P(X < 768) = P(Z < 0.30) \approx 0.617911[/tex].
[tex]\begin{aligned}P(624 < X < 768) &= P(X < 768) - P(X < 624) \\ &\approx 0.982136 - 0.617911\\ &\approx 0.60 \end{aligned}[/tex].
