SOMEONE PLEASE HELP A GIRL OUT BRAINLIEST! The diameters of Breaburn apples are normally distributed with a mean of 8.2cm and a standard deviation of 2.8cm. If you had a shipment of 2000 apples, how many apples would have a diameter: a) smaller than 5.9cm? b) larger than 10.2cm?

Respuesta :

Answer:

a. 0.2061*2000 apples = 412.20 or about 412 apples.

b. 0.2389*2000 apples = 477.8 or about 478 apples.

Step-by-step explanation:

We have to find the corresponding probabilities using the standard normal table (or also we can use statistics software for doing this).

The parameters for this normal distribution are (without using units):

The mean, [tex] \\ \mu = 8.2[/tex].

The standard deviation, [tex] \\ \sigma=2.8[/tex]

We need to use z-scores, given by:

[tex] \\ z = \frac{x-\mu}{\sigma}[/tex]

For a diameter smaller than 5.9cm, we have:

[tex] \\ z = \frac{5.9-8.2}{2.8}[/tex]

The value for z:

[tex] \\ z = -0.82142 \approx -0.82[/tex]

Then, find the cumulative probability for this z, P(z<-0.82), using the standard normal table (for this, use -0.8 (in right column), and -0.02 in the first row of the table).

Then, P(z<-0.82) = 0.2061

How many apples would have a diameter: a) smaller than 5.9cm?

The answer is 0.2061*2000 apples = 412.20 or about 412 apples.

How many apples would have a diameter: b) larger than 10.2cm?

We can proceed similarly as before. Calculate z, find the cumulative probability using the standard normal table.

[tex] \\ z = \frac{10.2-8.2}{2.8}[/tex]

[tex] \\ z = 0.71428 \approx 0.71[/tex]

Then, P(z<0.71) = 0.7611

But we are asked for P(z>0.71), then (because of the symmetry of the normal distribution):

P(z>0.71) = 1 - P(z<0.71)

P(z>0.71) = 1 - 0.7611

P(z>0.71) = 0.2389

The answer is 0.2389*2000 apples = 477.8 or about 478 apples.