Question

The weights of items produced by a company are normally distributed with a mean of 4.5 ounces and a standard deviation of 0.3 ounce. a. What is the probability that a randomly selected item from the production will weigh at least 4.14 ounces? b. What percentage of the items weighs between 4.8 and 5.04 ounces? c. Determine the minimum weight of the heaviest 5% of all items produced. d. If 27,875 items of the entire production weigh at least 5.01 ounces, how many items have been produced?

Answer 1

Answer:

The right solution is:

(a) 0.8849

(b) 12.28%

(c) 4.9935

(d) 28004

Step-by-step explanation:

Given:

Mean,

= 4.5

Standard deviation,

= 0.3

(a)

P(x > 4.14)

As we know,

⇒ $z = \frac{4.14-4.5}{0.3}$

      $=-1.20$

then,

⇒ $P(z>-1.20) = P(z<1.20)$

                          $=0.8849$

(b)

P(4.8 < x < 5.04)

= $P(\frac{4.8-4.5}{0.3} < \frac{x-\mu}{\sigma} < \frac{5.04-4.5}{0.3} )$

= $P(1<z<1.80)$

= $P(z<1.80)-P(z<1)$

= $0.9641 -0.8413$

= $0.1228$

or,

= $12.28$ (%)

(c)

P(x > x) = 0.05

z value will be,

= 1.645

⇒ $1.645 = \frac{x - 4.5}{0.3}$

         $x = 4.9935$

(d)

P(x < 5.01)

⇒ $z = \frac{x- \mu}{\sigma}$

      $=\frac{5.01-4.5}{0.3}$

      $=1.7$

P(z < 1.70) = 0.9554

⇒ $n = \frac{27875}{0.9954}$

       $=28004$