Respuesta :
Answer:
Step-by-step explanation:
Since the heights of students at a college are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = heights of students
µ = mean height
σ = standard deviation
From the information given,
µ = 175 cm
σ = 6 cm
The probability that the height of a student is less than 163 cm is expressed as
P(x < 163)
For x = 163
z = (163 - 175)/6 = - 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.023
Therefore, the expected number of students with heights lesser than 163 cm is
1000 × 0.023 = 23 students
Using the normal distribution, it is found that the number of students with heights less than 163 cm should be expected to be of 23.
In a normal distribution with mean and standard deviation , 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:
- Mean of 175 cm, thus [tex]\mu = 175[/tex].
- Standard deviation of 6 cm, thus [tex]\sigma = 6[/tex].
The proportion with heights less than 163 cm is the p-value of Z when X = 163, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{163 - 175}{6}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a p-value of 0.023.
Out of 1000 students:
0.023 x 1000 = 23
The number of students with heights less than 163 cm should be expected to be of 23.
A similar problem is given at https://brainly.com/question/24663213