Write a summary paragraph based on your findings on the distribution of the sale

Question

Write a summary paragraph based on your findings on the distribution of the sales data.
Based on the data, what is the probability that the sales amoung the telephone associates is A) more than 52k? B) Less than 32k? $45001-$50000 $50001 $55000 More $45001 $500 $50001-$550 24 Column1 Histogram Mean 45001.77433 Standard Error 145.3952348 44908.44306 45600 Standard Devia 2944.029121 Sample Varian 8667307.467 0.09007649 0.13394864 17422.60689 Minimum 34577.39311 Median Mode 100 Kurtosis Range > $35000 $35001 $40000 $40001-$45000 $45001-$50000 $0001 $55000 SALARY BAMS 52000 18450727.47 410 52000 Smallest(1) 34577.39311 Confidence Les 285.8152001 Maximum Largest(1) 0 Type here to search PrtScn Home di) F4

Explanation / Answer

Let X = sales among the telephone associates.

We assume X ~ N(µ, 2), where µ = 45001.77 and = 2944.03

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then,

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

Now, to work out the solutions,

Part (A)

Probability the sales among the telephone associates is more than 52k

= P(X > 52000)

= P[Z > {(52000 – 45001.77)/2944.03}] [vide (2) above]

= P(Z > 2.3771)

= 0.0087 [using Excel Function on Normal Distribution] ANSWER

Part (B)

Probability the sales among the telephone associates is less than 32k

= P(X < 32000)

= P[Z < {(32000 – 45001.77)/2944.03}] [vide (2) above]

= P(Z < - 4.4163)

= 0.000005 [using Excel Function on Normal Distribution] ANSWER

Part (C)

Probability the sales among the telephone associates is between 44k and 50k

= P(44000 < X < 50000)

= P[{(44000 – 45001.77)/2944.03} < Z < {(50000 – 45001.77)/2944.03}] [vide (2) above]

= P(- 0.3403 < Z < 1.6978)

= P(Z < 1.6978) - P(Z < - 0.3403)

= 0.9552 – 0.3669

= 0.5883 [using Excel Function on Normal Distribution] ANSWER

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