This article contains solved Statistics assignments for students covering topics like probability, binomial distribution, Poisson distribution, normal distribution, and hypothesis testing. It also includes the university name, course code, and student ID.
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Running Head: STATISTICS ASSIGNMENT Statistics Assignment Name of the Student Name of the University Student ID
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2STATISTICS ASSIGNMENT Answer 1 Probability of defective items (p) = 0.03 Number of randomly selected pieces (n) = 6 Let X be the number of defective parts. X ~ Bin (6, 0.03) a.P[X=0]=(6 0)(0.03)0(1−0.03)6=0.83 b.P[X=2]=(6 2)(0.03)2(1−0.03)4=0.01 Answer 2 Let X be the number of students having a credit card balance greater than $ 3,500. The probability of students having credit card balance greater than $ 3,500 (p) = 0.25. The number of students selected randomly (n) = 10. Therefore, X ~ Bin (10, 0.25) a.P[X=6]=(10 6)(0.25)6(1−0.25)4=0.02 b.P[X≥4] ¿1−P[X≤4] ¿1−(P[X=0]+P[X=1]+P[X=2]+P[X=3]) ¿1−[((10 0)(0.25)0(1−0.25)10 )+((10 1)(0.25)1(1−0.25)9 )+ ((10 2)(0.25)2(1−0.25)8 )+((10 3)(0.25)3(1−0.25)7 )]
3STATISTICS ASSIGNMENT ¿1−(0.06+0.19+0.28+0.25) ¿0.22 Answer 3 The number of employees (n) = 150. The expected number of accidents outside of work per year = 5 The probability of accidents per year (p) =5 150=0.033 Let X be the number of accidents per year. X ~ Bin (150, 0.033) a.P[X=0]=(150 0)(0.033)0(1−0.03)150=0.01 b.P[X≥4] ¿1−P[X≤4] ¿1−(P[X=0]+P[X=1]+P[X=2]+P[X=3]) ¿1−[((150 0)(0.033)0(1−0.33)150 )+((150 1)(0.033)1(1−0.033)149 )+ ((150 2)(0.033)2(1−0.033)148 )+((150 3)( ¿1−(0.01+0.03+0.08+0.14) ¿0.73 Answer 4 Let X be the number of small business failures per month. The average number of small business failures in a month (λ) = 30.
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4STATISTICS ASSIGNMENT Therefore, X ~ Poisson (30). P[X=4]=e−30×304 4!=3.16×e−9 Answer 5 Let X be the number of arrivals in a minute. The average number of arrivals in a minute (λ) = 7 Therefore, X ~ Poisson (7) a.P[X=4]=e−7×74 4!=0.09 b.P[X≥4] ¿1−P[X<4] ¿1−(P[X=0]+P[X=1]+P[X=2]+P[X=3]) ¿1−(e−7×70 0!+e−7×71 1!+e−7×72 3!+e−7×74 4!) ¿1−(0.0009+0.0064+0.0223+0.0521) ¿0.91 Answer 6 Let X be the remuneration of an information system manager. The average hourly compensation rate (μ) = $ 80. Standard deviation of the compensation (σ) = $ 15.50 X ~ Normal (80, 240.25)
5STATISTICS ASSIGNMENT a.P[80<X<85] ¿P[X<85]−P[X<80] ¿P(X−80 15.50<85−80 15.50)−P(X−80 15.50<80−80 15.50) ¿Φ(0.32)−Φ(0) ¿0.6255−0.5 ¿0.1255 b.P[X<55] ¿P(X−80 15.50<55−80 15.50) ¿Φ(−1.61) ¿0.0533 Answer 7 Average debt of a person (μ) = $ 25, 100. Standard Deviation of the debt (σ) = $ 3,750 Let X be the debt of a person. Therefore, X ~ Normal (μ,σ2) a.P[X>19000] ¿1−P[X<19000] ¿1−P(X−25100 3750<19000−25100 3750) ¿1−Φ(−1.63) ¿1−0.0519
6STATISTICS ASSIGNMENT ¿0.9481 b.P[X<15000] ¿P(X−25100 3750<15000−25100 3750) ¿Φ(−2.69) ¿0.003 c.P[22000<X<28000] ¿P[X<28000]−P[X<22000] ¿P(X−25100 3750<28000−25100 3750)−P(X−25100 3750<22000−25100 3750) ¿Φ(0.77)−Φ(−0.83) ¿0.7803−0.2042 ¿0.5761 Answer 8 Let X be the elapsed time of a user in making payments. Average time (μ) = 2 minutes. Standard Deviation of the time (σ) = 0.35 minutes. Size of the random sample drawn (n) = 30. P[2.3<X<3.8] ¿P[X<3.8]−P[X<2.3]
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7STATISTICS ASSIGNMENT ¿P (X−2 3.5 √30 <3.8−2 3.5 √30)−P (X−2 3.5 √30 <2.3−2 3.5 √30) ¿Φ(2.82)−Φ(0.47) ¿0.9976−0.6806 ¿0.3169 Answer 9 Average weight of cereal package = 32 grams Standard deviation of the weight = 1.22 grams Sample size = 30 P[X<31.3] ¿P (X−32 1.22 √30 <31.3−32 1.22 √30) ¿Φ(−3.14) ¿0.0008 Answer 10 A pharmaceutical company claims that one of their products is 99% effective. Therefore, population proportion (P) = 0.99 Sample size (n) = 575.
8STATISTICS ASSIGNMENT Obtained good results = 500. Sample proportion =500 575=0.87. Null Hypothesis (H0): P = 0.99 Alternate Hypothesis (HA): P≠0.99 Level of significance = 0.05 Test Statistic is given by: t=p−P √p×(1−p) n =0.87−0.99 √0.87×(1−0.87) 575 =−29.0247 The tabulated value of z at 0.05 percent level of significance is 1.96. The absolute value of the observed value of the test statistic is 29.02, which is greater than the tabulated value of the test statistic. Thus, the null hypothesis is rejected. Thus, it can be said that the claim made by the pharmaceutical company is not true.