Answer to Question #173471 in Statistics and Probability for ANJU JAYACHANDRAN

Question

4. (a) Five unbiased dice were thrown 96 times and the number of times 4, 5 or 6 was 

obtained, is given in the following table: (5) 

 

No. of dice 

showing 4, 

5 or 6 

0 1 2 3 4 5

Frequency 1 10 24 35 18 8

At 5% level of significance test whether this data comes from a binomial distribution. 

You may like to use the following values. 

 [( 10 05. ) 11 07. ,

2

x5 = 05.0( ) 12 59. ,

2

x6 = 05.0( ) 14 07. .] 2

x7 = 

 (b) The yield (in kg) of 100 plots in the form of grouped frequency distribution is given 

below: (5) 

 

Yield 

(kg) 

Frequency

0-20 6 

20-40 21 

40-60 35 

60-80 30 

80-100 8 

(i) Estimate the number of plots with an yield of 

(A)40 to 80 kg 

(B) 10 to 70 kg 

 (ii) Find the mean and standard deviation of yield.

Answer 1

Let H$_o$: The given data comes from Binomal distribution

Total number of dice is thrown N=96 times

Probability of obtaining 4,5 or 6 $p=\dfrac{1}{2}$

$q=1-p=1-\dfrac{1}{2}=\dfrac{1}{2}$

Therefore The expected frequency of obtaining 4,5 or 6 in throw of 5 dice 96 rimes as-

$N(r)=96\times ^5C_rp^rq^{1-r}$

The table for Chi-square is given by,-

The value of $\chi^2=\sum \dfrac{(O_i-E_i)^2}{E_i}=14.01$

The tabulate value of $\chi^2$ at 5% level of significance is 14.07.

Conclusion: As the calculated value of $\chi^2$ is less than the tabulated value So H_o is accepted i.e Data comes from binomal distribution.

(b) The table for given data is-

(i) Number of plots for yiled of 40-80 kg is 75.

Number of plots for yiled of 10-70 kg is 74.

(ii) Mean yiled $\bar{x}=\dfrac{\sum x_i.f_i}{\sum f_i}=\dfrac{5800}{100}=58$

Standard deviation $\sigma=\sqrt{\dfrac{(x_i-\bar{x})^2}{100}}=\sqrt{\dfrac{4320} {100}}=\sqrt{43.20}=6.57$