Normal
μ 2.23
σ 0.28
xi P(X<=xi)
1.61 0.0134
1.76 0.0466
2.39 0.7161
2.53 0.8580
P(X<=xi) xi
0.10 1.8712
0.20 1.9943
0.30 2.0832
0.40 2.1591
Normal
μ 2.23
σ 0.21
xi P(X<=xi)
1.61 0.0016
1.76 0.0126
2.39 0.7769
2.53 0.9234
P(X<=xi) xi
0.10 1.9609
0.20 2.0533
0.30 2.1199
0.40 2.1768

The mean weight for a part made using a new production process is 2.23 pounds. Assume that a normal distribution applies and that the standard deviation is 0.28 pounds. Based on this paragraph of text, use the correct excel output above to answer the following question.
The probability is .80 that weight for a part (in pounds) made using the new production process will be between two values equidistant from the mean. Find the upper value (in pounds).

a) 2.5888 b) 2.3009 c) 2.2832 d) 2.4991 e) None of the answers is correct

Option A

The upper value of the two values (equidistant from the mean) that cover 80% of the distribution is 2.5888 pounds

Step-by-step explanation:

This is a normal distribution problem as the probability distribution presented in the tables before the question rightly attest to.

Taking any of the values in the table and finding the percentage of weights of new parts less than or equal to that weight, using the z-score and the normal distribution table, we obtain the corresponding probabilities on the table.

This is the biggest proof that this distribution is indeed a normal distribution with

Mean = μ = 2.23 points

Standard deviation = σ = 0.28

The probability is 0.80 that the weight for a part (in pounds) made using the new production process will be between two values equidistant from the mean

Let the two values equidistant from the mean be x' and x"

The given probability is

P(x' < x < x") = 0.80

Let their respective z-scores be z' and z"

P(x' < x < x") = P(z' < z < z") = 0.80

Noting that the mean for a normal distribution has a z-score of 0, and weights above and below the mean have a proportion of 50%, that is, 0.5 each.

The lower limit of the weights we require will be more than 10% of the distribution and the upper limit of the weights we require will be more than 90% of the distribution.

These proportions (10% and 90%) are exactly each 40% from the mean (equidistant) and cover 80% of the total distribution equidistant from the mean.

So, to find the upper limit now,

P(x < x") = 90% = 0.90

P(x < x") = P(z < z") = 0.90

Using the normal distribution tables, the z-score that corresponds to 0.90 is 1.282.

z-score is given as the weight minus the mean weight divided by the standard deviation.

z" = (x" - μ)/σ

1.282 = (x" - 2.23)/0.28

x" = 2.23 + (1.282 × 0.28)

x" = 2.58896 pounds ≈ 2.5888 pounds (from the options provided)

Hence, the upper value of the two values equidistant from the mean that cover 80% of the distribution is 2.5888 pounds.

Hope this Helps!!!

i think it would be 42cm