Iff the hight of 1000 students are normally distributed with the mean of 68.5 inches and standard deviation of 2.7 inches how many of these students would you axpect to have hight? . Between 67.5 and 71.0 inches exclusive? . Greater then or equal to 74.0 inches?

Step-by-step explanation:

Let x be the random variable representing the height of the students. Since it is normally distributed and the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 68.5

σ = 2.7

the probability that a student's height is between 67.5 and 71.0 inches exclusive is expressed as

P(67.5 < x < 71.0)

For x > 67.5,

z = (67.5 - 68.5)/2.7 = - 0.37

Looking at the normal distribution table, the probability corresponding to the z score is 0.36

For x < 71.0

z = (71.0 - 68.5)/2.7 = 0.93

Looking at the normal distribution table, the probability corresponding to the z score is 0.82

Therefore,

P(67.5 < x < 71.0) = 0.82 - 0.36 = 0.46

The number of students whose height is between 67.5 and 71.0 inches is

0.46 × 1000 = 460 students

The probability that a student's height is greater than or equal to 74.0 inches is expressed as P(x ≥ 74)

P(x ≥ 74) = 1 - P(x < 74)

For x < 74,

z = (74 - 68.5)/2.7 = 2.04

Looking at the normal distribution table, the probability corresponding to the z score is 0.98

P(x ≥ 74) = 1 - 0.98 = 0.02

The number of students whose height are greater than or equal to 74.0 inches is

0.02 × 1000 = 20 students

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