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madam [21]
3 years ago
8

The average cost of tuition, room and board at small private liberal arts colleges is reported to be $8,500 per term with a stan

dard deviation of $1,200, but a financial administrator believes that the average cost is MORE. A study conducted using 350 small liberal arts colleges showed that the average cost per term is $8,745. Let alpha = 0.05. What is the critical z-value for this test?
Mathematics
1 answer:
Svetlanka [38]3 years ago
3 0

Answer:

z=3.8196

Step-by-step explanation:

-We notice that this is a normal distribution problem.

-Given the mean is $8500, standard deviation is $1200 , and n=350, the critical z-value using alpha=0.05 is calculated as:

z=\frac{x-\mu_o}{\sigma/\sqrt{n}}\\\\=\frac{8745-8500}{1200\sqrt{350}}\\\\\\=3.8196

Hence, the test statitic is z=3.8196

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Solve the equation -2( 1 - 5v)
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Help if u do both I’ll mark Brainlynest
amm1812

Answer:

See step-by-step explanation

Step-by-step explanation:

16. C, <AOB = 20

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6 0
3 years ago
Read 2 more answers
The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean 1263 and a stan
DENIUS [597]

Answer:

(a) The 29th percentile for the number of chocolate chips in a bag is 1198.65.

(b) The number of chocolate chips in a bag that make up the middle 95​% of bags are [1146, 1380].

(c) The inter-quartile range of the number of chocolate chips in a bag of chocolate chip​ cookies is 157.83.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of chocolate chips in a bag of chocolate chip cookies.

The random variable <em>X</em> is normally distributed with mean, <em>μ </em>= 1263 and a standard deviation, <em>σ </em>= 117.

(a)

Compute the 29th percentile for the number of chocolate chips in a bag as follows:

P (X < x) = 0.29

⇒ P (Z < z) = 0.29

The value of <em>z</em> for the above probability is, <em>z</em> = -0.55.

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sogma}\\-0.55=\frac{x-1263}{117}\\x=1263-(117\times 0.55)\\x=1198.65

Thus, the 29th percentile for the number of chocolate chips in a bag is 1198.65.

(b)

According to the Empirical rule 95% of the normally distributed data lies within 2 standard deviations of the mean.

P (μ - σ < X < μ + σ) = 0.95

P (1263 - 117 < X < 1263 + 117) = 0.95

P (1146 < X < 1380) = 0.95

Thus, the number of chocolate chips in a bag that make up the middle 95​% of bags are [1146, 1380].

(c)

The inter-quartile range of the normal distribution is:

IQR = 1.349 <em>σ</em>

Compute the inter-quartile range of the number of chocolate chips in a bag of chocolate chip​ cookies as follows:

IQR = 1.349 <em>σ</em>

      = 1.349 × 117

      = 157.833

Thus, the inter-quartile range of the number of chocolate chips in a bag of chocolate chip​ cookies is 157.83.

5 0
3 years ago
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