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frozen [14]
3 years ago
11

The IQs of 700 applicants to a certain college are approximately normally distributed with a mean of 115 and a standard deviatio

n of 11. If the college requires an IQ of at least 97, how many of these students will be rejected on this basis of IQ, regardless of their other qualificiations
Mathematics
1 answer:
Kobotan [32]3 years ago
3 0

Answer: 35

Step-by-step explanation:

Given : The IQs of 700 applicants to a certain college are approximately normally distributed with a mean of 115 and a standard deviation of 11.

i.e. \mu=115  and \sigma= 11

Let x denotes the IQs of applicants to college.

If the college requires an IQ of at least 97, then, the probability that students have IQ less than 97:-

P(x [By using z-table]

Number of students will be rejected on this basis of IQ = Total students x Probability of students have IQ less than 97

= 700 x 0.0505 = 35.35 ≈ 35

Hence, about 35 students will be rejected on this basis of IQ, regardless of their other qualifications .

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18. Determine the common difference, the fifth term, and the sum of the first 100 terms of the following sequence:
Ksenya-84 [330]

a_1=1,\ a_2=2.5,\ a_3=4,\ a_4=5.5,\ ...\\\\a_2-a_1=2.5-1=1.5\\a_3-a_2=4-2.5=1.5\\a_4-a_3=5.5-4=1.5\\a_{n+1}-a_n=1.5=constans\\\\\text{It's an arithmetic sequence with}\\a_1=1,\ \boxed{d=1.5}\\\\a_n=a_1+(n-1)d\to a_n=1+(n-1)(1.5)=1+1.5n-1.5\\\\a_n=1.5n-0.5\\\\a_5=1.5(5)-0.5=7.5-0.5=7\\\boxed{a_5=7}\\\\\text{The formula of a Sum of the First n Terms of an Arithmetric Sequence:}\\\\S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n\\\\\text{We have:}\\a_1=1,\ d=1.5,\ n=100\\\\\text{Substitute}

S_{100}=\dfrac{(2)(1)+(100-1)(1.5)}{2}\cdot100=\dfrac{2+(99)(1.5)}{1}\cdot50\\\\=(2+148.5)\cdot50=150.5\cdot50=7,525\\\\\boxed{S_{100}=7,525}\\\\Answer:\\the\ common\ difference:\ d=1.5\\the\ fifth\ term:\ a_5=7\\the\ sum\ of\ first\ 100\ terms:\ S_{100}=7,525

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3 years ago
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