Given a normal distribution of raw scores on an exam in which the -80 and sigma

Question

Given a normal distribution of raw scores on an exam in which the -80 and sigma 7.5, determine the following: mean Note: Before you attempt this problem, draw a curve for each component except b,h, and i. a) The percent of respondents who scored 60 or less b) The probability of locating a respondent who scored 60 or less e) The percent of respondents who scored between 80 and 90 d) The probability of locating a respondent who scored between 80 and 90 e) The percent of respondents who scored 85 or more The percent of respondents who scored between 85 and 95 x g) The percent of respondents who scored between 70 and 75 h) The probability of locating a respondent who scored either 70 or less or 90 or more The probability of obtaining three respondents who scored 90 or more. 1) II. In an impoverishe d urban area. individual income is normally distributed with a l Check Compatibility Mode

Explanation / Answer

Answer:

Given:

Normal distribution on raw scores in Exam

Mean = 80

SD = 7.5

a) The percentage of students who scored 60 or less

Using Excel function: NORM.DIST(60,80,7.5,FALSE) = 0.152% Ans.

b) The probability of locating the respondent who scored 60 or less

The Z value at score 60 is given by:

Z = (X – µ)/? = (60 – 80)/7.5 = -2.67

Now probability of locating the respondent who scored 60 or less is given by:

P(Z<= -2.67) , looking up the Z score in Z table, we get 1-0.962 = 0.38%

Alternatively,

Using Excel function: NORM.DIST(60,80,7.5,TRUE) = 0.383% Ans.

The percent of respondents who scored between 80 or 90

Using excel function, we have required percentage =

NORM.DIST(80,80,7.5,FALSE) – NORM.DIST(90,80,7.5,FALSE)

= 0.0532 – 0.0219 = 0.0313 or 3.13% Ans.

The probability of locating the respondent who scored between 80 and 90

Using excel function, we have required percentage =

NORM.DIST(90,80,7.5,TRUE) – NORM.DIST(80,80,7.5,TRUE)

= 0.9088 – 0.5000 = 0.4088 = 40.88% Ans.

e) The Percentage of respondents, who scored 85 or more

1 – NORM.DIST(85,80,7.5,TRUE) = 1 – 0.7475 = 0.2525 = 25.25% Ans.

f) The Percentage of respondents, who scored between 85 and 95 x

NORM.DIST(95,80,7.5,TRUE) - NORM.DIST(85,80,7.5,TRUE)

= 0.9772 – 0.7475 = 0.2297 = 22.97% Ans.

g) The Percentage of respondents, who scored between 70 and 75

NORM.DIST(75,80,7.5,TRUE) - NORM.DIST(70,80,7.5,TRUE)

= 0.2525 – 0.0912 = 0.1613 = 16.13% Ans.

h) The probability of locating a respondent who scored either 70 or less or 90 or more

NORM.DIST(70,80,7.5,TRUE) + 1 – NORM.DIST(90,80,7.5,TRUE)

= 0.0912 + (1-0.9088) = 0.1824 = 18.24% Ans

i) The probability of obtaining three respondents who scores 90 or more

Let Probability of locating respondent who scored 90 or more = P(A)

P(A) = 1 – NORM.DIST(90,80,7.5,TRUE) = 1-0.9088 = 0.0912

P(A | getting three respondents) = 0.0912*0.0912*0.0912 = 0.0007586 = 0.076% Ans.

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