In a large section of a statistics class, the points for the final exam are normally distributed, with a mean of 71 and a standard deviation of 7. Grades are assigned such that the top 10% receive A's, the next 20% received B's, the middle 40% receive C's, the next 20% receive D's, and the bottom 10% receive F's. Find the lowest score on the final exam that would qualify a student for an A, a B, a C, and a D.

The lowest score on the final exam that would qualify a student for an A is 80.

The lowest score on the final exam that would qualify a student for a B is 74.68.

The lowest score on the final exam that would qualify a student for a C is 67.33.

The lowest score on the final exam that would qualify a student for a D is 62.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mean of 71 and a standard deviation of 7.

This means that

Grades are assigned such that the top 10% receive A's, the next 20% received B's, the middle 40% receive C's, the next 20% receive D's, and the bottom 10% receive F's.

This means that:

90th percentile and above: A

70th percentile and below 90th: B

30th percentile to the 70th percentile: C

10th percentile to the 30th: D

Lowest score for an A:

Top 10% receive A, which means that the lowest score that would qualify a student for an A is the 100 - 10 = 90th percentile, which is X when Z has a pvalue of 0.9, so X when Z = 1.28.

The lowest score on the final exam that would qualify a student for an A is 80.

Lowest score for a B:

70th percentile, which is X when Z has a pvalue of 0.7, so X when Z = 0.525.

The lowest score on the final exam that would qualify a student for a B is 74.68.

Lowest score for a C:

30th percentile, which is X when Z has a pvalue of 0.3, so X when Z = -0.525.

The lowest score on the final exam that would qualify a student for a C is 67.33.

Lowest score for a D:

10th percentile, which is X when Z has a pvalue of 0.1, so X when Z = -1.28.

The lowest score on the final exam that would qualify a student for a D is 62.

The answer is in the image 

The answer is in the image 

The total nom of  code that can be used is equal to 5+3 = 8

F=ma

where F=force

m=mass

a=acceleration

Here,

F=4300

a=3.3m/s2

m=F/a

    =4300/3.3

    =1303.03kg

F=ma

where F=force

m=mass

a=acceleration

Here,

F=4300

a=3.3m/s2

m=F/a

    =4300/3.3

    =1303.03kg

Approximately it is aqual to 1300kg

The answer is in the image 

Salesperson will make 6% of 1800

=(6/100)*1800

=108

Salesperson will make $108 in $1800 sales

The solution is given in the image below

Here,

tip=18%of $32

tip=(18/100)*32

=0.18*32

=$5.76

Total payment=32+5.76=$37.76