Answer:
The mean of this distribution is approximately 3.96.
Step-by-step explanation:
Here's how to solve this problem using a normal distribution table.
Let [tex]z[/tex] be the
[tex]\displaystyle z = \frac{x - \mu}{\sigma}[/tex].
In this question, [tex]x = 5[/tex] and [tex]\sigma = 2[/tex]. The equation becomes
[tex]\displaystyle z = \frac{5 - \mu}{2}[/tex].
To solve for [tex]\mu[/tex], the mean of this distribution, the only thing that needs to be found is the value of [tex]z[/tex]. Since
The problem stated that [tex]P(X \le 5) = 69.85\% = 0.6985[/tex]. Hence, [tex]P(Z \le z) = 0.6985[/tex].
The problem is that the normal distribution tables list only the value of [tex]P(0 \le Z \le z)[/tex] for [tex]z \ge 0[/tex]. To estimate [tex]z[/tex] from [tex]P(Z \le z) = 0.6985[/tex], it would be necessary to find the appropriate
Since [tex]P(Z \le z) = 0.6985[/tex] and is greater than [tex]P(Z \le 0) = 0.50[/tex], [tex]z > 0[/tex]. As a result, [tex]P(Z \le z)[/tex] can be written as the sum of [tex]P(Z < 0)[/tex] and [tex]P(0 \le Z \le z)[/tex]. Besides, [tex]P(Z < 0) = P(Z \le 0) = 0.50[/tex]. As a result:
[tex]\begin{aligned}&P(Z \le z)\\ &= P(Z < 0) + P(0 \le Z \le z) \\ &= 0.50 + P(0 \le Z \le z)\end{aligned}[/tex].
Therefore:
[tex]\begin{aligned}&P(0 \le Z \le z) \\ &= P(Z \le z) - 0.50 \\&= 0.6985 - 0.50 \\&=0.1985 \end{aligned}[/tex].
Lookup [tex]0.1985[/tex] on a normal distribution table. The corresponding [tex]z[/tex]-score is [tex]0.52[/tex]. (In other words, [tex]P(0 \le Z \le 0.52) = 0.1985[/tex].)
Given that
Solve the equation [tex]\displaystyle z = \frac{x - \mu}{\sigma}[/tex] for the mean, [tex]\mu[/tex]:
[tex]\displaystyle 0.52 = \frac{5 - \mu}{2}[/tex].
[tex]\mu = 5 - 2 \times 0.52 = 3.96[/tex].
