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olasank [31]
2 years ago
4

In a large population of college-educated adults, the mean iq is 112 with standard deviation 25. suppose 300 adults from this po

pulation are randomly selected for a market research campaign. the probability that the sample mean iq is greater than 115 is:
Mathematics
1 answer:
valkas [14]2 years ago
8 0
Given:
μ = 112, population mean
σ = 25, population std. deviation
The randomly selected sample population of 300 (>30) is large enough for probability testing based on the normal distribution.

x = 115, random variable.
z-score:  
x = (x-μ)/σ = (115 - 112)/25 = 0.12

Let xs = sample mean.
From standard tables, obtain
P(xs<115) = 0.548

Therefore
P(xs>115) = 1 - 0.548 = 0.452 = 45% (approx)

Answer: 45%
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Assume that the number of apples is x and the number of oranges is y.

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we also know that the total amount spent is $12, therefore the first equation is as follows:
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2 years ago
a 12 foot tall building casts on 8 foot long long shadow. how long of a shadow will a 5 foot tall women have
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7 0
2 years ago
Use the normal approximation to the binomial distribution to answer this question. Fifteen percent of all students at a large un
Nataly_w [17]

Answer: 0.1289

Step-by-step explanation:

Given : The proportion of all students at a large university are absent on Mondays. : p=0.15

Sample size : n=12

Mean : \mu=np=12\times0.15=1.8

Standard deviation = \sigma=\sqrt{np(1-p)}

\Rightarrow\ \sigma=\sqrt{12(0.15)(1-0.15)}=1.23693168769\approx1.2369

Let x be a binomial variable.

Using the standard normal distribution table ,

P(x=4)=P(x\leq4)-P(x\leq3)              (1)

Z score fro normal distribution:-

z=\dfrac{x-\mu}{\sigma}

For x=4

z=\dfrac{4-1.8}{1.2369}\approx1.78

For x=3

z=\dfrac{3-1.8}{1.2369}\approx0.97

Then , from (1)

P(x=4)=P(z\leq1.78)-P(z\leq0.97)\\\\=0.962462-0.8339768\approx0.1289    

Hence, the probability that four students are absent = 0.1289

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