with the mean 1200 hrs and standard deviation number of bulbs in a lot of 2500 bulbs having the burning life.

i. move than 1300 hours.

ii. between 1050 and 1400 hours.

We have $ Z= \frac {X-m}{\sigma} $

Given, m=1200, $\sigma$ = 90, N= 2500

$ Z= \frac {X-1200}{90} -----------1 $

Case 1: more than 1300 hours

$ \therefore X = 1300 in \ equation\ 1$

$ Z= \frac{1300-1200}{90}$

Z= 1.111

$ P(Z \geqslant 1300)= area\ to \ the\ right\ of 1.111 $

= 0.5 - (Area between Z=0 and Z=1.111)

= 0.5 - 0.335

$ P(Z \geqslant 1300)= 0.1335 $

$ \therefore\ Number\ of \ bulb\ with \ burning\ life\ more\ than\ 1300\ hours $

$ = N \times p (Z \geqslant 1300)$

= 2500 $\times$ 0.1335

= 333.75

$\approx$ 334 bulbs

ii. between 1050 and 1400 hours

X= 1050 & X = 1400 IN 1

$ Z= \frac {1050-1200}{90} \ \ \ \ Z= \frac {1400-1200}{90}$

$ Z= -1.6667 \ & \ Z= 2.2222 $

$ P(1050 \leq Z \leq 1400 )= area \ between Z = -1.6667 \ and \ Z = 2.2222 $

= Area from (Z = 0 to 1.6667) + area from (Z = 0 to 2.2222)

= 0.4522 + 0.4869

= 0.9391

Number of bulbs with burning life between 1050 and 1400 hours

= N $\times (1050 \leq Z \leq 1400) = 2500 \times 0.9391 $

= 2347.75 $\approx$ 2348 bulbs