contestada

The mean weight of 500 college students is 70 kg and the standard deviation is 3 kg. Assuming that the weight is normally distributed, determine what percentage of students weigh over 75 kg

Respuesta :

JeanaShupp

Answer: 4.75%.

Step-by-step explanation:

Given: The mean weight [tex](\mu)[/tex] of 500 college students is 70 kg and the standard deviation[tex](\sigma)[/tex] is 3 kg.

Let X be the number of students weight over 75 kg.

If the weight is normally distributed, then the probability that students weigh over 75 kg:

[tex]P(X>75)=P(\dfrac{X-\mu}{\sigma}>\dfrac{75-70}{3})\\\\=P(z>\dfrac53)\\\\=P(z>1.67)\\\\=1-P(z<1.67)\\\\=1-0.9525\ \ \ [\text{By p-value table}]\\\\= 0.0475\\\\=4.75\%[/tex]

Hence, the  percentage of students weigh over 75 kg = 4.75%.

Using the normal distribution, it is found that 4.75% of students weigh over 75 kg.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It measures how many standard deviations the measure is from the mean.  
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

  • The mean is of 70 kg, hence [tex]\mu = 70[/tex].
  • The standard deviation is of 3 kg, hence [tex]\sigma = 3[/tex].

The proportion over 75 kg is 1 subtracted by the p-value of Z when X = 75, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{75 - 70}{3}[/tex]

[tex]Z = 1.67[/tex]

[tex]Z = 1.67[/tex] has a p-value of 0.9525.

1 - 0.9525 = 0.0475

0.0475 x 100% = 4.75%

4.75% of students weigh over 75 kg.

To learn more about the normal distribution, you can take a look at https://brainly.com/question/24663213

ACCESS MORE
ACCESS MORE
ACCESS MORE
ACCESS MORE

Otras preguntas