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natali 33 [55]
3 years ago
6

Scores on the Wechsler intelligence quotient (IQ) test are normally distributed with a mean score of 100 and a standard deviatio

n of 15 points. The US military has minimum enlistment standards at about an IQ score of 85. There have been two experiments with lowering this to 80 but in both cases these recruits could not master soldiering well enough to justify the costs. Based on IQ scores only, what percentage of the population does not meet US military enlistment standards?
Mathematics
1 answer:
rosijanka [135]3 years ago
5 0
15.87% of the population does not meet the minimum standards.

We use z-scores to find this probability:

z=(X-μ)/σ = (85-100)/15 = -15/15 = -1

Using a z-table (http://www.-table.com) we see that the area to the left of, less than, this is 0.1587.  This is 15.87%.
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ladessa [460]
Formulas are mathematical laws that help us solve problems.

Decimals are a type of number; graphs are displays of data; percentages are comparisons of a number to 100; and fractions are a type of number.
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2 years ago
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two friends natasha and tricia share a sum of money in the ratio 5:3 respectively if tricia share was $126.75,calculate the tota
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Answer:

253.48

Step-by-step explanation:

Hello this is answer

7 0
2 years ago
Solve the equation and check.<br> - 14 = x + 14
guajiro [1.7K]

The answer will be -28. Hope it helps

6 0
2 years ago
A survey of 700 adults from a certain region​ asked, "What do you buy from your mobile​ device?" The results indicated that 59​%
Kryger [21]

Answer:

a) the test statistic z = 1.891

the null hypothesis accepted at 95% level of significance

b) the critical values of 95% level of significance is zα =1.96

c) 95% of confidence intervals are  (0.523 ,0.596)

Step-by-step explanation:

A survey of 700 adults from a certain region

Given sample sizes n_{1} = 400 and n_{2} = 300

Proportion of mean p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52

<u>Null hypothesis H0</u> : assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

<u>Alternative hypothesis H1:</u>- p1 ≠ p2

a) The test statistic is

Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )}  } }

where p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}  

on calculation we get   p = 0.56    

now q =1-p = 1-0.56=0.44

Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )}  } }\\   =\frac{0.56-0.52}{\sqrt{0.56X0.44}(\frac{1}{400}+\frac{1}{300}   }

after calculation we get z = 1.891

b) The critical value at 95% confidence interval zα = 1.96 (from z-table)

The calculated z- value < the tabulated value

therefore the null hypothesis accepted

<u>conclusion</u>:-

assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

c) <u>95% confidence intervals</u>

The confidence intervals are P± 1.96(√PQ/n)

we know that = p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}

after calculation we get P = 0.56 and Q =1-P =0.44

Confidence intervals are ( P- 1.96(√PQ/n), P+ 1.96(√PQ/n))

now substitute values , we get

( 0.56- 1.96(√0.56X0.44/700), 0.56+ 1.96(0.56X0.44/700))

on simplification we get (0.523 ,0.596)

Therefore the population proportion (0.56) lies in between the 95% of <u>confidence intervals  (0.523 ,0.596)</u>

<u></u>

3 0
3 years ago
Multiplies to -1120 adds to 3
TEA [102]
With these, always write out the multiples first.

Start like this:
(assume one of the factors is negative)
1 and 1120
2 and 560
4 and 280
5 and 224
7 and 160
8 and 140
10 and 112
14 and 80
16 and 70
20 and 56
28 and 40
32 and 35

from those, the obvious choice is the one with a difference of three. In this case, 32 and 35, because -32 + 35 equals 3.
6 0
3 years ago
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