Question

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A company that manufactures bookcases finds that the average time it takes an employee to build a bookcase is 10 hours with a standard deviation of 2 hours. A random sample of 64 employees is taken. What is the likelihood that the sample mean will be 9 hours or more? The average grade point average (GPA) of undergraduate students in New York is normally distributed with a population mean of 2.5 and a population standard deviation of .5. Compute the following, showing all work:

(I) The percentage of students with GPA's between 1.3 and 1.8 is: (a) less than 5.6% (b) 5.7% (c) 5.9% (d) 6.2% (e) 6.3% (f) 6.6% (g) 7.3% (h) 7.5% i) 7.9% (j) more than 8%.

(II) The percentage of students with GPA's below 2.3 is:

(III) Above what GPA will the top 5% of the students be (i.e., compute the 95th percentile):

(IV) If a sample of 36 students is taken, what is the probability that the sample mean GPA will be between 2.60 and 2.75

4. At the end of the Halloween Festival, the organizers estimated that a family of participants spent in average of $45.00 with a standard deviation of $10.00. If 49 participants (49 = size of the sample) are selected randomly, what's the likelihood that their mean spent amount will be within $4 of the population mean? (mean +/- 4)

Answer 1

A company that manufactures bookcases finds that the average time it takes an employee to build a bookcase is 10 hours with a standard deviation of 2 hours. A random sample of 64 employees is taken. What is the likelihood that the sample mean will be 9 hours or more?

The average grade point average (GPA) of undergraduate students in New York is normally distributed with a population mean of 2.5 and a population standard deviation of .5. Compute the following, showing all work:

(I)

The percentage of students with GPA's between 1.3 and 1.8 is:    

Choice (g) 7.3%

(II)

The percentage of students with GPA's below 2.3 is:

= 34.5%

(III)

Above what GPA will the top 5% of the students be (i.e., compute the 95th percentile):    

= 3.32

(IV)

If a sample of 36 students is taken, what is the probability that the sample mean GPA will be between 2.60 and 2.75?  

= 0.1137

Explanation:

normally distributed

population mean, μ = 2.5

population standard deviation, σ = 0.5

Compute the following.

(I) The percentage of students with GPA's between 1.3 and 1.8 is:    

P(1.3<X<1.8)

Convert the the grades to z scores

P(1.3<X<1.8)=P(-2.4<z<-1.4)

=P(z<-1.4)-P(z<-2.4)

=0.08076-0.00820

= 0.07256

As a percentage

= 0.07256*100%

=7.256%

= 7.3% Choice (g) 7.3%

(II)

The percentage of students with GPA's below 2.3 is:

P(X<2.3)

P(X<2.3)=P(z<-0.4)

= 0.34458

As a percentage

=0.34458*100%

=34.458%

= 34.5%

(III)

Above what GPA will the top 5% of the students be (i.e., compute the 95th percentile):    

95th percentile

P(z<Z)=0.95

Z score for 95th percentile = 1.64

from the z score formula, z=X−μ​/σ

X=μ+z∗σ

=2.5+1.64*0.5

= 3.32

(IV)

If a sample of 36 students is taken, what is the probability that the sample mean GPA will be between 2.60 and 2.75?  

sample size, n = 36

P(2.60<X<2.75)

Use formula for the z score of the sampling of the mean

P(2.60<X<2.75)=P(1.2<z<3)

= P(z<3)-P(z<1.2)

= 0.99865-0.88493

= 0.11372

= 0.1137

4. At the end of the Halloween Festival, the organizers estimated that a family of participants spent in......

The likelihood that their mean spent amount will be with $4 of the population mean is = 0.99488

Explanation:

Given Data

Mean , μ = 45

Standard deviation , σ = 10

Sample size , n = 49

Required is  ±4

x1​ = Lower bound = μ - 4 = 45 - 4 = 41

x1​ = Upper bound = μ + 4 = 45 + 4 = 49

We first get the z score for the two values.

As z=x−μ *​​ (√ n)/σ

z1​ = Lower z score = x1​ - μ * (​​√ n)/σ

Substitute the values: x1​ = 41 , μ = 45 , σ = 10 and n =49

z1​ = (41 - 45) *(√ 49/)10=-2.8

From normal distribution tables 

z1​ = 0.00256

z2​ = Upper z score = x2​ - μ *(√ n)/σ

Substitute the values: x2​ = 49 , μ = 45 , σ = 10 and n = 49

z2​ = (49 - 45) * √ 49/10​​ = 2.8

From normal distribution table 

Therefore, z2​ = 0.99744 

We substrate the values value of z1​ from z2​

P(z1<z<z2)=0.99744 - 0.00256 = 0.99488 

Therefore, The likelihood that their mean spent amount will be with $4 of the population mean is = 0.99488