Mathematics, 18.03.2020 03:28 blevinssam274

The heights of students at a college are normally distributed with a mean of 175 cm and a standard deviation of 6 cm. One might expect in a sample of 1000 students that the number of students with heights less than 163 cm is:

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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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 175$ .Standard deviation of 6 cm, thus $\sigma = 6$ .

The proportion with heights less than 163 cm is the p-value of Z when X = 163, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{163 - 175}{6}$

Z = -2

Z = -2 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 link

Answer from: ella3714

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

Answer from: Quest

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

= $\sqrt{(9-6)^{2} +(12-6)^{2} } \\=\sqrt(45)=3\sqrt(5)$

side j'k'=distance between j' and k'

= $\sqrt{(3-2)^{2} +(2-4)^{2} } \\=sqrt(5)$

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl

Answer from: Quest

Aquadratic function is one of the form f(x)=ax2 + bx + c l, where a,b, and c are numbers with are not equal to zero.

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