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Answer: Z 0 and Z 1.63 0.4484: X Mean Standard Deviation

The document contains 7 problems involving calculations using the normal distribution. The problems include finding areas under the normal curve between given z-scores, determining standard scores, expected values and variances. The document also contains questions finding corresponding scores given the population mean and standard deviation.

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Sam Madronero
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346 views3 pages

Answer: Z 0 and Z 1.63 0.4484: X Mean Standard Deviation

The document contains 7 problems involving calculations using the normal distribution. The problems include finding areas under the normal curve between given z-scores, determining standard scores, expected values and variances. The document also contains questions finding corresponding scores given the population mean and standard deviation.

Uploaded by

Sam Madronero
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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1. Find the area under the normal curve between z = 0 and z = 1.63.

Answer: z = 0 and z = 1.63


= 0.4484

2. On a final examination in Mathematics, the mean was 76 and the standard deviation was 5.
Determine the standard score of a student who received a score of 88 assuming the scores are normally
distributed.
x −mean
¿
Answer: Standard Deviation

Mean = 76 88−76
¿
5
Standard Deviation = 5
12
¿
x = 88 5

= 2.4

3. What percent of the area under the normal curve is between z = - 1.00 and z = 1.00?

Answer: z = - 1.00 and z = 1.00


-1.00 = 0.3413 and 1.00 = 0.3413
= 0.3413 + 0.3413
= 0.6826
= 68.26 %
4. Find the area under the normal curve between z = 0 and z = - 1.78

Answer: z = 0 and z = -1.78


= 0.4625

5. Anthony Vincent tosses an unbiased coin. He receives Php 100 if a head appears and he pays Php 40 if
a tail appears. Find the expected value and the variance of his gain.

x P(x) xP(x) x2P(x)


-40 0.5 -20 800
100 0.5 50 5,000
= 30 = 5,800
Answer:

a. expected value = [xP(x)]


= 30
 The expected value is Php 30

b. variance of his gain = [x2P(x) - ([xP(x)])^2


= 5,800 - (30)^2
 The variance of his gain is Php 4,900
6. In a given normal distribution, the population mean is 60 and the population standard deviation is 3.2.
Find the corresponding score of 68

Answer: x −mean
¿
Standard Deviation
Mean = 60
68−60
¿
Standard Deviation = 3.2 3.2
x = 68 8
¿
3.2

= 2.5

7. In a given normal distribution, the population mean is 60 and the population standard deviation is 3.2.
Find the corresponding score of 56.

Answer: x −mean
¿
Standard Deviation
Mean = 60
56−60
¿
Standard Deviation = 3.2 3.2
x = 56 −4
¿
3.2

= -1.25

PS. NAKARAWAT AKON BAYAD NA


TWICE ALBUM (FEEL SPECIAL)

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