Question

solve step by step explanation ​

Answer 1

Answer:

p = 14

q = 29

Step-by-step explanation:

We are given the following measures of the diameters of the heads of 150 screws below;

Diameter (in mm)   No. of screws (f)     Revised CI            X            X * f

  32 - 36                       15                         31.5 - 36.5           34           510

   37 - 41                        17                         36.5 - 41.5           39           663

   42 - 46                       p                          41.5 - 46.5           44           44*p

   47 - 51                        25                        46.5 - 51.5           49          1225

   52 - 56                       q                          51.5 - 56.5           54          54*q

   57 - 61                        20                        56.5 - 61.5           59         1180

   62 - 66                       30                        61.5 - 66.5           64        1920    

                          ∑f = 107+p+q                                              ∑Xf=5498+44p+54q

Here, we have revised Class Interval(CI) of  Diameter(in mm) because in the original CI upper limit of previous class is not same with the lower limit of next class. So, for doing this we have subtracted 0.5 from lower limit and added 0.5 to the upper limit.

We are given that total no. of screws = 150

          107 + p + q = 150

             p + q = 43

                  q = 43 - p ---------- [Equation 1]

Mean formula is given by = $\frac{\sum Xf}{\sum f}$

              51.2 mm      =   $\frac{5498+44p+54q}{107+p+q}$

             51.2(107 + p + q) = 5498 + 44p + 54q

             51.2(107 + p + 43 - p) = 5498 + 44p + 54(43 - p) {using equation 1}

              51.2 * 150 = 5498 + 44p + 2322 - 54p

                   10p = 7820 - 7680

                   p = 140 / 10 = 14

Now, putting value of p in equation 1 we get, q = 43 - 14 = 29 .