Question

The scores of fourth grade students on a mathematics achievement test follow a normal distribution with a mean of 75 and standard deviation of 4.
a) What is the probability that a single student randomly chosen form all those taking the test scores 80 or higher?
b) What is the probability that the sample mean score of 64 randomly selected student is 80 or higher?
c) What is the probability the sample mean score of 64 randomly selected students is between 74 and 76?

Answer 1

Answer:

The appropriate solution is:

(a) 0.1056

(b) 0

(c) 0.9544

Step-by-step explanation:

The given values are:

Mean,

$\mu = 75$

Standard deviation,

$\sigma = 4$

(a)

⇒  $P(x>80)=1-(x$

                     $=1-P[\frac{x-\mu}{\sigma}$

                     $=1-P[\frac{x-\mu}{\sigma}$

                     $=1-P(z$

By using the table, we get

                     $=1-0.8944$

                     $=0.1056$

(b)

According to the question, the values are:

$n$ = 64

$\mu_\bar{x}$ = 75

Now,

⇒  $\sigma_\bar{x}$ = $\frac{\sigma}{\sqrt{n} }$

         = $\frac{4}{\sqrt{64} }$

         = $\frac{4}{8}$

         = $0.5$

⇒  $P(\bar {x} >80 ) = 1 - P(\bar x$

                      $=1 - P[\frac{(\bar x-\mu_\bar x)}{\sigma \bar x} < \frac{80-75}{0.5} ]$

                      $=1-P(z$

By using the table, we get

                      $=1-1$

                      $=0$

(c)

As we know,

⇒ $\sigma_\bar x$ = $\frac{\sigma}{\sqrt{n} }$

        = $\frac{4}{\sqrt{64} }$

        = $\frac{4}{8}$

        = $0.5$

then,

= $P(74< \bar x$

= $P[\frac{74-75}{0.5} < \frac{\bar x-\mu \bar x}{\sigma \bar x} < \frac{76-75}{0.5} ]$

= $P(-2$

= $P(z$

By using the table, we get

= $0.9772-0.0228$

= $0.9544$