It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below. Weight (in kg) Less than38 38-40 40-42 42-44 44 46 46-48 48-50 50-52 Total(n) Frequency (f) 0 3-0=3 5-3=2 9-5=4 14-9=5 28-14=14 32-28=4 35-Read more
It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below.
Weight (in kg) Less than38 38-40 40-42 42-44 44 46 46-48 48-50 50-52 Total(n)
Frequency (f) 0 3-0=3 5-3=2 9-5=4 14-9=5 28-14=14 32-28=4 35-32=3 35
Cumulative
frequency 0 3 5 9 14 28 32 35
The cumulative frequency just greater than n/2 (i.e. 35/2 = 17.5) is 28, belonging to class intervals 46- 48.
Median class = 46 – 48
lower class limit (l) of median class = 46
Cumulative frequency (cf) of class preceding median class = 14
Class size (h) = 2
Median = l + ((n/2 – cf)/f) × h = 46 + ((17.5 – 14)/14) × 2 = 46 + 0.5 = 46.5
Therefore, median of this data is 46.5.
Hence, the value of median is verified.
There are a total of 5 days. Shyam can go to the shop in 5 ways and Ekta can go to the shop in 5 ways. Therefore, total number of outcomes = 5 × 5 = 25 (i) They can reach on the same day in 5 ways. ie, (t, t), (w, w), (th, th), (f, f),(s, s) P(both will reach on same day) = 5/25 = 1/5 (ii) They canRead more
There are a total of 5 days. Shyam can go to the shop in 5 ways and Ekta can go to the shop in 5 ways.
Therefore, total number of outcomes = 5 × 5 = 25
(i) They can reach on the same day in 5 ways.
ie, (t, t), (w, w), (th, th), (f, f),(s, s)
P(both will reach on same day) = 5/25 = 1/5
(ii) They can reach on consecutive days in these 8 ways- (t, w), (w, th), (th, f), (f, s), (w,t), (th, w), (f, th), (s, f).
Therefore, P (both will reach on consecutive days) = 8/25
P (both will reach on same day) = 1/5(From (i)]
P (both will reach on different days) = 1 – 1/5 = 4/5
Write the following in decimal form and say what kind of decimal expansion each has :3/13
(iv) 3/13 = 0. 230769230769230769…, Recurring & Non-terminating
(iv) 3/13 = 0. 230769230769230769…, Recurring & Non-terminating
See lessWrite the following in decimal form and say what kind of decimal expansion each has :2/1.
(v) 2/11 = 0.181818…, Recurring & Non-terminating
(v) 2/11 = 0.181818…, Recurring & Non-terminating
See lessWrite the following in decimal form and say what kind of decimal expansion each has : 329/400.
(vi) 329/400 = 0.8225, Terminating
(vi) 329/400 = 0.8225, Terminating
See lessDuring the medical check-up of 35 students of a class, their weights were recorded as follows:
It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below. Weight (in kg) Less than38 38-40 40-42 42-44 44 46 46-48 48-50 50-52 Total(n) Frequency (f) 0 3-0=3 5-3=2 9-5=4 14-9=5 28-14=14 32-28=4 35-Read more
It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below.
Weight (in kg) Less than38 38-40 40-42 42-44 44 46 46-48 48-50 50-52 Total(n)
Frequency (f) 0 3-0=3 5-3=2 9-5=4 14-9=5 28-14=14 32-28=4 35-32=3 35
Cumulative
frequency 0 3 5 9 14 28 32 35
The cumulative frequency just greater than n/2 (i.e. 35/2 = 17.5) is 28, belonging to class intervals 46- 48.
See lessMedian class = 46 – 48
lower class limit (l) of median class = 46
Cumulative frequency (cf) of class preceding median class = 14
Class size (h) = 2
Median = l + ((n/2 – cf)/f) × h = 46 + ((17.5 – 14)/14) × 2 = 46 + 0.5 = 46.5
Therefore, median of this data is 46.5.
Hence, the value of median is verified.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?
There are a total of 5 days. Shyam can go to the shop in 5 ways and Ekta can go to the shop in 5 ways. Therefore, total number of outcomes = 5 × 5 = 25 (i) They can reach on the same day in 5 ways. ie, (t, t), (w, w), (th, th), (f, f),(s, s) P(both will reach on same day) = 5/25 = 1/5 (ii) They canRead more
There are a total of 5 days. Shyam can go to the shop in 5 ways and Ekta can go to the shop in 5 ways.
See lessTherefore, total number of outcomes = 5 × 5 = 25
(i) They can reach on the same day in 5 ways.
ie, (t, t), (w, w), (th, th), (f, f),(s, s)
P(both will reach on same day) = 5/25 = 1/5
(ii) They can reach on consecutive days in these 8 ways- (t, w), (w, th), (th, f), (f, s), (w,t), (th, w), (f, th), (s, f).
Therefore, P (both will reach on consecutive days) = 8/25
P (both will reach on same day) = 1/5(From (i)]
P (both will reach on different days) = 1 – 1/5 = 4/5